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ELECTRICAL  ENGINEERING 
FIRST  COURSE 


McGraw-Hill  BookCompany 


Electrical  World         The  Engineering  and  Mining  Journal 
Engineering  Record  Engineering  News 

Railway  Age  Gazette  American  Machinist 

Signal  Hngin<?er  ^American Engineer 

Electric  Railway  Journal  Coal  Age 

Metallurgical  and  Chemical  Engineering  Power 


ELECTRICAL 
ENGINEERING 

FIRST  COURSE 


BY 
ERNST  JULIUS  BERG,  Sc.  D. 

PROFESSOR  OP  ELECTRICAL  ENGINEERING 
UNION  COLLEGE,  SCHENECTADY,  N.  Y. 

AND 

WALTER  LYMAN  UPSON,  E.  E.,  M. 

ASSOCIATE   PROFESSOR  OF    ELECTRICAL    ENGINEERING 
UNION  COLLEGEJ|SCHENECTADY,  N.   Y. 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1916 


COPYRIGHT,  1916,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


THK    MAPI.*    PHKS8    YORK 


PREFACE 

A  text-book  in  electrical  engineering  emanating  from  Union 
College  may  be  the  occasion  of  some  surprise  to  those  who  have 
been  conversant  with  the  development  of  the  electrical  course 
in  that  institution.  The  authors  have,  it  is  true,  recorded  their 
objections  to  the  use  of  a  prescribed  text.  These  objections  still 
hold  good.  In  brief,  they  are,  first,  that  a  prescribed  text  tends 
to  take  the  life  out  of  the  class  room,  whether  the  course  be  con- 
ducted by  lectures  or  recitations,  and  second,  that  it  tends  to  take 
the  life  out  of  the  study  by  relieving  the  student  of  responsibility 
of  continued  effort. 

At  Union  College  the  fundamental  aim  is  that  the  student 
shall  first  comprehend,  and  then,  create.  Comprehension  comes 
through  directed  effort.  This,  the  student  acquires  readily  in 
the  laboratory,  but  in  the  class  room,  it  is  not  so  easy.  The 
recitation  falls  short  because  it  deals  with  the  individual  rather 
than  the  class.  The  lecture  fails  when  the  student  knows  he  can 
fall  back  upon  the  text-book.  The  fault,  however,  is  not  with 
the  text-book  itself,  but  with  the  use  that  is  made  of  it. 

Obviously,  then,  its  proper  use  is  as  a  means  of  directing  the 
student's  effort  toward  comprehension.  Indeed,  it  should  com- 
pel effort,  not  in  order  to  make  up  for  an  author's  failure  to  ex- 
press himself  clearly,  but  in  order  that  the  ideas  shall  sink  in  and 
make  permanent  impressions  on  the  mind.  The  book  should, 
therefore,  be  so  constructed  and  used  that  it  shall  be  an  additional 
aid  to  the  student  in  creating  his  own  expression  of  the  ideas 
with  which  he  is  brought  into  contact  in  the  lecture,  the  recitation 
and  the  laboratory.  It  is  desirable  that  fundamental  ideas  shall 
become  fixed  and  clear  in  the  student's  mind  as  soon  as  possible, 
thus  leaving  him  in  a  position  to  exert  his  full  mental  effort  on 
that  which  is  more  advanced. 

As  he  progresses,  he  should  acquire,  more  and  more,  the  power 
of  self-direction,  that  is,  the  power  to  create  or  construct  his  own 
ideals.  Creative  work  finds  its  primary  impulse  in  imitation. 
The  student  should  have  before  him,  at  the  outset,  a  model, 
which  he  is  faithfully  to  copy. 


vi  PREFACE 

In  attempting  to  embody  these  principles  in  the  present 
volume,  the  authors  have  sought  to  maintain  a  harmonious  inter- 
relationship between  the  book  and  the  class  room.  The  lectures 
which  form  the  basis  of  the  book  were  first  delivered  eight 
years  ago.  In  that,  and  subsequent  years,  students  have  had  to 
rely  for  assistance  upon  their  own  notes  and  on  help  received 
individually  from  instructors.  Many  of  the  problems  assigned 
are  now  worked  out  completely  or  in  part  in  the  text.  They 
thus  cease  to  be  available  for  assignment,  but  the  ideas  contained 
in  them  have  been  extended  to  form  new  problems  whose  solutions 
will  be  obtained  only  after  study  of  the  problems  solved. 

These  new  problems  have  generally  remained  unsolved,  in  the 
past,  owing  to  lack  of  available  time.  It  is  believed  that  they 
may  now  be  carried  through  with  fair  completeness  and,  indeed, 
that  many  other  suggestions  coming  from  them  may  be  followed. 

It  is  the  belief  of  the  authors  that  no  book  on  Electrical  Engi- 
neering can  now  be  produced  which  does  not  bear  testimony  to  the 
pioneer  work  of  such  writers  as  Fleming,  Silvanus  Thompson, 
Bedell  and  Crehore,  Steinmetz  and  McAllister. 

In  addition,  the  authors  desire  to  acknowledge  their  indebted- 
ness to  Dr.  A.  S.  McAllister  who  has  critically  gone  over  the 
manuscript,  to  Mr.  N.  S.  Diamant  for  suggestions  relating  to 
material  contained  in  the  earlier  portions  of  the  book,  and  to  Mr. 
E.  S.  Lee  for  assistance  in  reading  the  proof. 


CONTENTS 

PAGE 

PREFACE    v 

CHAPTER  I 

UNITS -      1 

Development  of  Ohm's  Law — Two  Resistances  in  Parallel — 
General  Solution  of  a  Network  by  KIRCHOFF'S  Laws — Effects  of 
Current  in  a  Wire — Power. 

CHAPTER  II 


FORM  OF  WORK 9 

Object  of  Problems — Summary. 

CHAPTER  III 

* 

MAGNETISM 14 

Cylindrical  Poles — Flat  Poles — Magnets  as  Commonly  Used  in 
Meters — Energy  Density  in  a  Field — Limits  of  Pole  Intensity — 
The  Magnetic  Cycle — Permeability. 

CHAPTER  IV 

PRINCIPLE  OF  THE  ELECTRIC  MOTOR 20 

Determinations  of  Magnetic  Intensity — Magnetic  Intensity  at  Any 
Point  along  the  Axis  of  a  Coil — Magnetic  Intensity  in  the  Center 
of  a  Long  Coil — Application  of  Magnetic  Formulae  to  Instruments. 

CHAPTER  V 

DESIGN  OF  A  LIFTING  MAGNET 26 

CHAPTER  VI 

GENERATION  OF  ELECTROMOTIVE  FORCE  IN  A  DYNAMO 29 

E.m.f.  Waves  in  Fields  that  are  not  Uniform — E.m.f.  Wave  when 
the  Coil  is  Wound  on  an  Iron  Core — Additional  Problems  for 
the  Determination  of  E.m.f.  Waves. 

CHAPTER  VII 

INDUCTANCE 33 

The  Rate  of  Energy  Supply  or  Power  Equation — Starting  and 
Stopping  Current  in  an  Inductive  Circuit. 

vii 


viii  CONTENTS 

CHAPTER  VIII 

PAGE 

ALTERNATING  CURRENTS 38 

Average  Value  of  a  Sine  Wave — Effective  Value  of  a  Sine  Wave. 

CHAPTER  IX 

DIRECT-CURRENT  GENERATORS 44 

Homopolar  Generators — Direct-current  Machines  with  Commu- 
tators— Types  of  Direct-current  Commutator  Machines — Arma- 
ture Reaction — Characteristics  of  Direct-current  Generators — 
Numerical  Application. 

CHAPTER  X 

A  STUDY  OF  THE  DESIGN  OF  A  DIRECT-CURRENT  GENERATOR  ....  55 
Flux  Calculation — The  Magnetic  Circuit — Area  of  Flux  Path 
through  Teeth — Area  of  Flux  Path  through  Gap — Areas  of 
Armature  Core,  Pole  Core  and  Yoke — Materials — No-load  and 
Full-load  Saturation  Curves — Armature  Reaction — The  Shunt 
Field  Winding — The  Series  Field  Winding — Armature  Resistance 
— Brush  Resistance — Series  Field  Resistance — Commutator  and 
Brushes — Flux  Distribution  around  the  Armature — Losses  and 
Efficiency — Copper  Losses — Core  Loss — Summary  of  Losses, 
Output  and  Efficiency — Temperature  Rise. 

CHAPTER  XI 

ELECTRICAL  CONSTANTS   OF  A  DIRECT-CURRENT  GENERATOR  HAVING 

COMMUTATING  POLES  AND  COMPENSATING  WINDING      74 

Commutation. 

CHAPTER  XII 

DIRECT-CURRENT  GENERATORS  IN  PARALLEL  AND  SERIES 88 

Direct-current  Generators  in  Series — The  Three-wire  System — 
Boosters. 

CHAPTER  XIII 

DIRECT-CURRENT  MOTORS 97 

Types  of  Direct-current  Motors — Speed  Characteristics  of  Direct- 
current  Motors — Power  and  Torque — Torque  Characteristics. 

CHAPTER  XIV 
THEORY  OF  THE  BALLISTIC  GALVANOMETER 102 

CHAPTER  XV 

VECTOR  REPRESENTATION  OF  ALTERNATING-CURRENT  WAVES      .    .    .   105 
Use  of  the  Symbol  j — Circuit  of  Resistance  in  Series  with  an 
Inductive  Impedance — Impedances  in  Parallel. 

./    '1 


CONTENTS  ix 

CHAPTER  XVI 

PAGE 

THE  SYMBOLIC  METHOD  IN  TRANSMISSION  LINE  CALCULATION   ...    113 
Addition — Multiplication — Power — Average  Value  of  Power  dur- 
ing a   Period — Power  Factor — Transmission  Line  Calculation — 
Power  of  Generator. 

CHAPTER  XVII 

CONSTANT   POTENTIAL — CONSTANT  CURRENT  TRANSFORMATION  .    .    .   120 
CHAPTER  XVIII 

CAPACITY  AND  CAPACITY  REACTANCE 123 

Condenser — Expression  of  Condensive  Impedance — Circuit  Con- 
taining Resistance,  Inductance  and  Capacity  in  Series — Resonance. 

CHAPTER  XIX 

PARALLEL  CIRCUITS 129 

Transmission  Line  Supplying  Power  to  Parallel  Loads — Approxi- 
mate Transmission  Line  Calculation. 

* 
CHAPTER  XX 

DISTORTED  WAVES — RESONANCE   EFFECTS 133 

E.m.f.  Which  Causes  Distorted  Current  waves. 

CHAPTER  XXI 

CONSTANT  POTENTIAL — CONSTANT  CURRENT  TRANSFORMATION  (Con- 
tinued from  Chapter  XVII) 140 

Power  and  Wattless  Components  of  Volt-amperes. 

CHAPTER  XXII 

THEORY  AND  USE  OF  THE  WATTMETER 146 

Wattmeter  Connections. 

CHAPTER  XXIII 

SIMPLE  PROBLEMS  IN  ELECTRO-STATICS 152 

Potential — Intensity — Capacity  of  a  Sphere — Potential  Gradient — 
Capacity  of  a  Spherical  Concentric  Condenser — The  Capacity  of 
a  Concentric  Cylinder — Capacity  of  Two  Parallel  Plates  so  Large 
that  the  Effects  of  their  Edges  may  be  Neglected — Capacity  of 
a  Transmission  Line — Capacity  of  a  Three-phase  Cable — Induc- 
tance of  a  Concentric  Cable — Inductance  of  a  Transmission  Line. 

CHAPTER  XXIV 
DISTRIBUTED  INDUCTANCE  AND  CAPACITY  ....   162 


x  CONTENTS 

CHAPTER  XXV 

PAGE 

NOTES  ON  THE  MATHEMATICS  OF  COMPLEX  QUANTITIES 169 

Representation  of  Complex  Quantities — Addition  of  Two  Complex 
Quantities — Multiplication  of  Two  Complex  Quantities — Division 
of  Two  Complex  Quantities — Multiplication — Involution  and  Evo- 
lution— The  Roots  of  a  Complex  Quantity — Exponential  Repre- 
sentation of  Complex  Quantities — Differentiation  of  a  Complex 
Number  or  Vector — Logarithm  of  a  Complex  Number  or  Vector. 

CHAPTER  XXVI 

THE  TRANSFORMER ; 174 

The  Transformer  Diagram — Equivalent  Transformer  Circuit — 
Example  of  Transformer  Calculation — Approximate  Method  of 
Determining  the  Regulation,  Efficiency  and  Power  Factor  of 
Transformers. 

CHAPTER  XXVII 

HYSTERESIS  AND  EDDY-CURRENT  LOSSES 186 

Hysteresis  Loss — Eddy-current  Loss. 

CHAPTER  XXVIII 

WAVE  DISTORTION  IN  TRANSFORMERS 189 

Effect  of  Hysteresis — Dependence  of  Core  Loss  on  the  Shape  of  the 
E.m.f.  Wave. 

CHAPTER  XXIX 

DISTORTED  WAVES 196 

Application  of  FOURIER'S  Theorem  to  Wave  Analysis. 

CHAPTER  XXX 

MECHANICAL  STRESSES  IN  TRANSFORMERS   . 202 

Determination  of  the  Leakage  Inductance  of  the  Primary  and 
Secondary — Calculation  of  Stresses. 

CHAPTER  XXXI 

GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN 209 

Type — Efficiency — Losses — B  and  V — Hysteresis  and  Eddy- 
current  Loss  per  Cubic  Inch — Magnetizing  Current — Number  of 
Turns,  Total  Flux,  Area  and  Length  of  Magnetic  Circuit — Resist- 
ance, Length  of  Mean  Turn,  Total  Length  and  Size  of  Windings — 
Per  Cent.  Magnetizing  Current  and  Core  Loss — Efficiency — 
Regulation— Heating— Weight  and  Cost  of  Material. 


CONTENTS  xi 

CHAPTER  XXXII 

PAGE 

COMBINATIONS  IN  MULTIPHASE  TRANSFORMER  SYSTEMS   ......   226 

The  Three-phase  System — Voltage  Waves  in  Three-phase,  Four- 
wire  System — Three-phase,  Y-connected  Transformers — Three- 
phase  A-connected  Transformers — Voltage  Waves  with  Y-con- 
nected Transformers — Three-phase  Transformers — Shell-type 
Three-phase  Transformers — Open  Delta  Transformer  Connection 
— T-connection  of  Transformers — Rating  of  T-connected  Trans- 
formers— Two-phase,  Three-phase  Transformation — rAuto-trans- 
formers — Compensators  for  Two-phase,  Three-phase  Transfor- 
mation— Dissimilar  Transformers  in  Series — Dissimilar  Trans- 
formers in  Parallel — Three-phase  Connection  for  Dissimilar 
Transformers. 

CHAPTER  XXXIII 

ALTERNATORS 246 

One-,  Two-  and  Three-phase  Connection — Voltage  to  Neutral — 
Rating  of  Alternators. 

CHAPTER  XXIV 

ARMATURE  REACTION 252 

CHAPTER  XXXV 

CHARACTERISTICS  OF  ALTERNATORS  WITH  DEFINITE  POLES     ....   258 
CHAPTER  XXXVI 

APPROXIMATE    DETERMINATION   OF   THE    SELF-INDUCTION   OR  LOCAL 

MAGNETIC  LEAKAGE  REACTANCE  OF  AN  ALTERNATOR 270 

CHAPTER  XXXVII 

ARMATURE  REACTION  IN  MULTIPHASE  MACHINES 278 

Effect  of  Distributed  Winding  on  the  Armature  Reaction. 

CHAPTER  XXXVIII 

HUNTING 283 

CHAPTER  XXXIX 

STUDY  OF  THE  DESIGN  CONSTANTS  OF  ALTERNATORS 289 

General  Constants — Slot  Dimensions — Flux  Determination — Air 
Gap — Teeth  Flux  Density — Armature  Length — Armature  Resist- 
ance— Magnetic  Circuit  Dimensions — The  Main  Field  Magneto- 
motive Force — Losses  and  Efficiency — Temperature  Rise — 
Regulation. 


xii  CONTENTS 

CHAPTER  XL 

PAGE 

SHORT-CIRCUIT  OF  ALTERNATORS 301 

Stresses  on  End-connections  of  the  Armature  Coils — Multiphase 
Short-circuits— Armature  Reaction — Electromotive  Force  and 
Current  Induced  in  the  Field  Windings. 

CHAPTER  XLI 

SYNCHRONOUS  MOTORS 324 

Synchronous  Motor  Equations. 

CHAPTER  XLII 

INDUCTION  MOTORS    .   .   .   ; 342 

The  Rotary  Field — Theory  of  Operation — Case  1.  Armature  at 
Standstill — Case  2.  Armature  at  about  Half-speed — Motor  As- 
sumed Y-connected — Motor  Assumed  A-connected — Motor  and 
Transmission  Line — Motor  with  Auto-transformer. 

CHAPTER  XLIII 

STUDY  OF  THE  DESIGN  CONSTANTS  OF  AN  INDUCTION  MOTOR    .    .    .  365 
Air  Gap— Rotor  Diameter — Stator  Slots  per  Pole — Slot  and  Tooth 
Dimensions — Main     Flux — Stator     Length — Rotor     Slots — Slot 
and  Tooth  Dimensions — Rotor  Secondary  Resistance — Leakage 
Reactance. 

CHAPTER  XLIV 

ROTARY  OR  SYNCHRONOUS  CONVERTERS 383 

Voltage  and  Current  Ratios— Voltage  Control— Heating  of  the 
Armature— Voltage  Control— Numerical  Application — Voltage 
Control  by  Use  of  the  "Split  Pole" — Transformer  Connections  for 
Rotary  Converters — Synchronous  Condensers. 

CHAPTER  XLV 

SINGLE-PHASE  ALTERNATING-CURRENT  MOTORS 400 

The  Series  Motor — Compensated  Series  Motor — Repulsion  Motor. 

.411 


ELECTRICAL  ENGINEERING 


CHAPTER  I 
UNITS 

As  it  is  assumed  that  the  student  has  had  an  elementary 
course  in  Physics,  it  seems  feasible  to  omit  herein  the  definition 
of  the  fundamental  mechanical  and  electrical  units.  However, 
before  taking  up  the  electrical  engineering  problems,  it  is  essential 
that  a  review  be  made  of  the  chapters  in  physics  relating  to  these 
units. 

The  student  should  be  able  to  present,  not  only  by  means  of 
equations,  but  in  words  —  for  this  is  far  more  important  —  the 
relations  between  force,  work,  energy*,  power,  torque,  etc. 

In  regard  to  electrical  units  it  is  assumed  that  he  is  already 
familiar  with  such  terms  as  "  current"  and  "electromotive  force" 
and  appreciates  that  .  .  . 

Current  is  analogous  to  water  flowing.  The  absolute  unit  of 
current  is  the  abampere.  The  practical  unit  is  the  ampere.  One 
abampere  is  10  amperes. 

Quantity,  likewise,  is  analogous  to  water  at  rest.  The  practical 
unit  of  quantity  is  the  coulomb,  which  is  the  amount  of  electricity 
involved  when  1  amp.  flows  throughout  1  sec.,  or,  1  amp.  sec. 

Difference  of  potential  is  analogous  to  pressure-difference  and  is 
the  electromotive  force  which  causes  current  to  flow  in  a  circuit. 

The  absolute  unit  of  potential-difference  is  the  abvolt.  The 
practical  unit  is  the  volt,  which  is  108  abvolts. 

Resistance  is  that  property  of  the  material  of  a  circuit  which  im- 
pedes the  flow  of  electricity.  The  absolute  unit  of  resistance  is 
the  abohm.  The  practical  unit  is  the  ohm,  which  is  109  abohms. 
Resistance  depends  on  material  and  temperature.  With  constant 
temperature, 


- 


2  ELECTRICAL  ENGINEERING 

where 

R  =  resistance  of  a  given  conductor, 

p  =  specific  resistance  or  resistivity  of  the  material, 

Z  =  length,  and  A  =  area  of  cross-section,  of  the  conductor. 

Specific  resistance  or  resistivity,  is  the  resistance  of  a  unit  cube  of 
"any  material  taken  between  opposite  faces. ! 

In  practice  it  is  sometimes  convenient  to  use  the  resistance  of  a 
wire  1  ft.  long  and  0.001  in.  in  diameter  as  the  unit  of  resistivity. 
This  unit  is  called  the  circular-mil-foot. 

In  problems  involving  resistance,  it  is  frequently  convenient  to 
use  the  reciprocal  of  resistance,  known  as  the  conductance. 

G  =  o>  where  G  is  the  conductance  of  a  circuit  of  resistance  R. 
K 

Likewise  the  reciprocal  of  resistivity,  called  conductivity,  is  often 
used. 

The  resistance  of  a  wire  at  any  temperature  t,  when  its  resistance 
at  any  other  temperature  is  known  can  be  calculated  by  the 
following  equation 

Rt  =  Rt.ll  +  atl(t  -  t,)} 
When  ti  =  0°C.  then  Rt  =  R0(l  +  aj) 

where  Rt  is  the  required  resistance  at  any  temperature,  t,  R0  in 
this  case  is  the  resistance  at  0°,  and  a  is  a  constant,  called  the 
temperature  coefficient. 

For  copper,  a0  =  0.004  (approximately)  when  t  is  given  in 
Centigrade  degrees. 

At  any  other  temperature  the  value  of  a  is: 

1 


234.5  +  t 

where  t  is  the  temperature  in  degrees  C. 

Since  a  depends  upon  the  temperature,  in  all  calculations  in- 
volving a  its  value  is  calculated  for  that  temperature  at  which  the 
resistance  is  known. 

Knowing  the  resistance  Ri  at  a  temperature  ti  the  resistance 
Rz  at  temperature  tz  is  thus  accurately  determined  from  the 
following  relation: 


234.5  +  t,       234.5  + 


UNITS  3 

TABLE  I 

Table  I  gives  approximately  the  temperature  coefficients  and  resistivities 
in  ohms  per  centimeter  cube  of  some  of  the  more  common  electrical  con- 
ductors at  ordinary  temperature. 


Conductor 

Temp,  coefficient  a 

Resistivity 

Aluminium                                  

0.0042 

2.9  X  10~6 

Carbon 

—  0  00052 

720  X  10~« 

Copper 

0.004 

1.6  X  10~6 

German  silver                             

0.00027 

20.9  X  10~6 

Iron 

0.0046 

9  7  X  10~6 

Nickel                     

0.0062 

12.4  X  10~« 

Platinum                                             .  .  . 

0.0036 

9  X  10-« 

Silver            .          

0.004 

1.5  X  10~6 

Tungsten  

0.005 

5  X  10-« 

Development  of  Ohm's  Law. — According  to  OHM'S  law  the 
current  in  a  circuit  at  any  instant  is  equal  to  the  potential  differ- 
ence divided  by  the  resistance,  or, 

7  =  R' 

Obviously,  where  a  number  of  resistances  are  in  series,  the 
total  resistance  is  the  sum  of  the  individual  resistances,  or, 

Rtotal    ~    2r    =   TI  +  TZ  +  7*3  -f-    .      .      . 

Two  Resistances  in  Parallel. — To  find  the  total  current  7,  and 
the  currents  7i,  Iz  in  the  resistances  r\  and  r2,  when  a  potential 
difference  E  is  applied  (Fig.  1).  / 

By  OHM'S  law, 

Tjl  Tjl 

r  Hi  £j 


and 


FIG.  1. 


E       E 


To  find  a  single  resistance,  r0,  which  shall  be  the  equivalent  of 
ri  and  r2  in  parallel,  evidently 


' 


4  ELECTRICAL  ENGINEERING 

Whence, 


ra  = 


r2 


Having  two  resistances  in  parallel,  in  series  with  a  third  resist- 
ance (Fig.  2),  to  find  the  combined  resistance.     Let  the  combined 

resistance  of  r*i  and  r2  be  r0.     Then  r0  =  -       — 

T\  -J-  7*2 

The  condition  is,  then,  that  of  two  re- 


i—  :  —  Sr  —  i  r      sistances  r0  and  r3  in  series  and  the  total 
j>*\      r   >i3 

n     resistance  R  =  r0  +  r3. 


+  J  • 

Hence 


r 

*  /  -  -  - 

-L        "•""        yv 


FIG.  2.  72       r0  +  r3 

To  find  /i  and  /2. 
It  is  evident  that  /  =  73. 

Knowing  73  and  r3,  we  may  at  once  determine  E$  which  is  the 
potential  difference,  or  drop,  across  r3.  Thus,  by  OHM'S  law, 

Ez  =  /3r3. 

It  is  evident  that  the  potential  difference  E0,  across  r\  and  T*  is 
E  —  E$. 

'      T       _    E°.     T        -    ^ 

-  Tt>  /2  ~   r2 

General  Solution  of  a  Network  by  Kirchoff's  Laws. — In  cir- 
cuits or  networks  of  a  more  complicated  nature  in  which  the 
resistances  and  electromotive  forces  are  known,  the  currents  in 
the  various  branches  may  be  calculated  by  the  application  of 
KIRCHOFF'S  laws  which  may  be  stated  as  follows: 

Law  I. — The  algebraic  sum  of  all  the  currents  flowing  toward 
a  branch  point  is  equal  to  zero.  «_ 

Law  II. — The  algebraic  sum  of  all  the 
e.m.fs.  acting  around  a  closed  circuit  is 
equal  to  the  sum  of  the  products,  ri, 
around  the  mesh.  Or  the  impressed 
e.m.f.  is  equal  to  the  sum  of  all  e.m.fs. 
consumed  by  the  resistances. 

For  example,  let  the  circuit  be  as 
shown  in  Fig.  3  where  arrows  represent  arbitrarily  chosen  direc- 
tions of  current.  For  the  points  A,  B,  C,  D,  applying  Law  I, 
equations  may  be  written: 


UNITS 


A. 
B. 
C. 
D. 


-    i  - 


0 

is  -  i*  -  t'4  =  0 
ii  +  12  -  is  =  0 
i4  +  *B  -  i  =  0. 


(1) 

(2) 
(3) 
(4) 


Applying  Law  II,  where  the  short  arrow  represents  the  direc- 
tion of  the  e.m.f.,  to  the  meshes  (a)  e,  r3,  r4,  (b)  e,  ri,  r5,  (c)  nf  r», 
r3,  (d)  r2,  r5,  r4,  always  keeping  an  arbitrarily  chosen  counter- 
clockwise direction,  we  have, 

(a)  ri  +  r3i8  +  r4i4  =  e  (5) 

(b)  ri  +  nil  +  rbib  =  e  (6) 

(c)  nil  -  r,i2  -  r3i3  =  0  (7) 

(d)  r2iz  +  r6i8  -  r4i4  =  0  (8) 

There  is  one  extra  equation  in  each  group  as  there  are  only 
six  unknown  quantities,  i,  ii,  i2,  is,  i4,  is. 

In  calculating  the  resistance  of  more  or  less  complex  circuits 
it  is  helpful  to  remember  that  current  does  not  flow  between 
points  of  the  same  potential. 

If,  in  Fig.  3,  there  is  no  difference  of  potential  between  points 
B  and  C  there  will  be  no  current  in  the  branch  r2. 

PROBLEMS 

Problem  1. — If  the  resistivity  (resistance  of  a  cubic  centimeter  between 
parallel  faces  at  0°C.)  of  copper  is  1.6  X  10~6  ohm,  (a)  show  that  the  resist- 
ance of  an  inch  cube  of  copper  is  0.63  X  10~6  ohm;  (b)  show  that  if  the 
temperature  coefficient,  a.  =  0.004,  the  resistance  of  a  centimeter  cube  at 

20°C.  is  1.73  X  10~6  ohm;  (c)  show  that  __ B 

the   temperature  coefficient  per   degree 
Fahrenheit  is  0.0022. 


_ 


FIG.  4. 


FIG.  5. 


Problem  2. — If  a  wire  be  connected  across  the  terminals  of  a  source  of 
constant  e.m.f.,  a  current  will  flow.  Will  this  current  increase,  decrease,  or 
remain  constant  as  time  goes  on,  and  why? 

Problem  3. — Deduce  the  equation  for  the  equivalent  resistance  of  three 
resistances  connected  in  parallel. 

Problem  4. — Find  the  line  current  7,  and  the  voltage  across  r3  in  the 
circuit,  shown  in  Fig.  4.  E  =  100  volts,  r:  =  1,  r2  =  2,  r3  =  3. 

Problem  5. — Let  the  outline  of  a  cube,  Fig.  5,  consist  of  resistances,  each 


6  ELECTRICAL  ENGINEERING 

edge  being  1  ohm.  Prove  that  the  total  resistance  between  A  and  B  is 
^f  2  ohm;  between  A  and  C  is  %  ohm;  between  A  and  D  is  %  ohm. 

Effects  of  Current  in  a  Wire. — When  a  current  is  set  up  in  a 
wire  three  effects  may  be  noted,  namely:  (1)  the  wire  gets  warm, 
(2)  a  compass  needle  placed  near  the  wire  is  deflected,  and  (3) 
when  the  voltage  is  high  enough  bits  of  paper  may  be  attracted. 

The  amount  of  energy  delivered  through  the  wire  does  not  bear 
a  relation  to  any  one  of  these  effects,  but  if  the  second  and  third 
effects  are  multiplied  together,  or,  as  commonly  expressed,  if 
the  strength  of  the  magnetic  and  electric  fields  are  multiplied  to- 
gether the  product  is  a  value  which  is  proportional  to  the  amount 
of  energy  transmitted  through  the  wire  per  second,  or  to  the 
power.  Thus  we  may  write, 

P  =  kei 

where  P  is  the  power  and  k  is  a  constant,  k  is  unity  when  e, 
which  is  proportional  to  the  strength  of  the  electric  field  is  ex- 
pressed in  volts,  it  which  is  proportional  to  the  strength  of  the 
magnetic  field  is  expressed  in  amperes,  and  P  is  in  watts. 

The  first  effect,  that  is,  the  production  of  heat  is  due  to  con- 
sumption of  energy  in  the  wire  due  to  its  resistance.  The  second 
effect  is  due  to  the  setting  up  of  a  magnetic  field  about  the  wire 
by  the  current.  The  third  effect  is  due  to  the  setting  up  of  an 
electric  or  electro-static  field  in  the  region  about  the  wire  by  the 
difference  of  potential  between  the  wire  and  other  points  in  space. 

Power. — In  a  given  circuit,  then, 

P  =  El  =  IE  X  /  =  PR 

in  which  E  is  the  total  e.m.f.,  I  the  current,  and  R  the  total  re- 
sistance of  the  circuit. 

This  relation,  known  as  JOULE'S  law,  is  very  important,  as  it 
shows  that  the  power  is  proportional  to  the  square  of  the  current 
strength  and  to  the  first  power  of  the  resistance. 

The  heat  developed  by  this  power  depends  upon  the  duration  of 
the  current,  and  is  expressed  in  joules.  Thus,  heat  energy  = 
Elt  =  I2Rt  joules,  where  E  is  in  volts,  7  in  amperes,  and  t  in 
seconds  (the  current  and  voltage  being  assumed  constant  during 
time  t). 

r* 

In  general,  the  energy  converted  to  heat  is  W  =  I    i2rdt. 

Jti 

Problem   6. — Prove   that   if   the   current   is   represented   by   equation 

i  —  I  sin  ut 


UNITS  7 

T 
the  energy  per  cycle  is  W  =  72r  -^  where  T  is  the  time  of  a  complete  cycle. 

4 

W     Pr 
The  average  power  is  then  -7=-  =  -=-• 

j.         z 

(/I2          /32\ 
~o~  +  ~o~)r  when 

i  =  /i  sin  co£  +  /3  sin  (3«J  +  a). 

Heat  Units.  —  The  practical  heat  units  most  frequently  dealt 
with  are  the  British  thermal  unit  (B.t.u.),  and  the  large  and  small 
calories  (C.  and  c.). 

One  B.t.u.  is  the  energy  required  to  raise  the  temperature  of  1 
Ib.  of  water  1°F. 

1  B.t.u.  =  1.055  kw.  sec. 

One  large  calorie  is  the  energy  required  to  raise  the  temperature 
of  1  kg.  of  water  1°C. 

1  C.  =  4.2  kw.  sec. 

One  small  calorie  is  the  energy  required  to  raise  the  temperature 
of  1  gram  of  water  1°C. 

1  c.  =  0.0042  kw.  sec. 

Problem  8.  —  A  16-cp.  lamp  which  consuntes  3  watts  per  cp.  is  immersed 
in  a  quart  of  water  at  20°C.  Assuming  no  loss  of  heat,  (a)  what  will  the 
temperature  of  the  water  be  after  2  min.  ?  (6)  How  long  would  it  take  to 
evaporate  the  water? 

Solution.—  (a)  Temp,  will  be  20°  +  °C.  rise. 

°C.rise  =  ^W^  Xqt.inlkg. 

2 
kw.  sec.  =  X  16  X  2  X  60  =  5.76 


qt.  per  kg.  =  1.057 

f\   7  A 

.'.  °C.  rise  =  -~-  X  1.057  =  1.45. 

Temp,  after  2  min.  =  21.45°C. 

(6)  Time  to  evaporate  =  time  to  raise  .to  boiling  +  time  required  to 
furnish  latent  heat  of  vaporization. 
Time  required  to  boil  1  qt.  =  time  to  raise  1  qt.  1°  X  (100°  -  20°) 

=  ~±  X  80  =  110.3  min. 


Time  required  to  evaporate  =  calories  required  to  evaporate  -*•  calories  per 

min.  supplied  by  lamp. 


=  742  min. 
.'.  Total  time   required  =  110.3  +  742  =  852.3  min.  =  14  hr.  12  min. 


8  ELECTRICAL  ENGINEERING 

Problem  9. — Transform  problem  8  into  °F.  and  B.t.u. 

Problem  10. — If  electric  energy  costs  lOc.  per  kw.  hr.,  how  much  would 
it  cos,t  to  prepare  a  hot  bath  by  electric  means,  if  the  bath  required  50  gal. 
of  water  raised  in  temperature  by  50°F.  ? 

Solution.— Cost  =  kw.  hr.,  X  $0.10 

_  kw.  sec.  _  kw.  sec,  to  raise  1  gal.  1°  X  50  X  50 
3600  3600 

1  gal.  weighs  approx.  8.4  Ib. 
/.  kw.  sec.  to  raise  1  gal.  1°  =  8.4  X  1.055  =  8.86 

8.86  X  2500 

'  '  kw'  hr'  =         3600      "  =  6'15 
Cost  =  6.15  X  0.10  =  $0.615. 

Problem,  11. — Four  car  heaters  each  take  4  amp.  at  125  volts.  Find  the 
cost  per  10-hr,  day  at  lOc  per  kw.  hr.,  to  operate  them  on  a  500-volt  circuit, 
(a)  when  they  are  connected  in  series,  (6)  when  they  are  connected  in  parallel. 

Answer.— In  series,  $2.00;  in  parallel,  $32.00. 

Problem  12. — If  the  car  contains  3000  cu.  ft.  and  is  insulated  against  loss 
of  heat,  how  much  time  is  required  for  a  rise  in  temperature  of  20°C.  when 
the  heaters  of  problem  11  are  connected  (a)  in  series,  (6)  in  parallel? 

Answer. — In  series,  12  min.  44  sec.;  in  parallel,  48  sec. 

NOTE. — Specific  heat  of  air  at  constant  volume  =  0.167. 


CHAPTER  II 
FORM  OF  WORK 

In  order  that  students  may  gain  the  greatest  possible  advantage 
from  pursuing  the  course  of  study,  it  has  been  thought  best  to 
include  in  the  body  of  the  book,  at  this  point,  a  brief  statement 
of  the  procedure  which  the  student  should  adopt  in  the  working 
out  of  the  problems.  He  is  urged  to  familiarize  himself  with  the 
method,  and  to  follow  it  rigidly  until,  in  so  doing,  he  has  thor- 
oughly acquired  the  habit  of  careful  and  accurate  work. 

Object  of  Problems. — Problems  are  almost  universally  con- 
sidered to  be  indispensable  in  any  engineering  course.  Their 
function  is  similar  in  many  respects  to  that  of  laboratory  experi- 
ments. They  illustrate  the  theory.  In  this  respect  problems 
may  be  divided  into  two  groups,  namely: 

(a)  Those  in  which  the  general  equation  is  applied  to  a  definite 
concrete  case,  and 

(b)  Those  in  which  the  general  equation  is  investigated  for  the 
purpose  of  finding  out  the  whole  range  of  definite  value  which 
may  be  obtained  from  one  variable  by  assigning  definite  values 
to  one  or  more  other  variables. 

As  an  illustration  of  the  first  group,  we  will  take  the  following 
example : 

Problem  13. — Ten  arc  lamps,  in  series,  are  used  to  light  a  certain  building. 
They  require  6.6  amp.,  and  the  potential-difference  (drop)  across  each  lamp 
is  80  volts.  Current  is  supplied  from  a  power  house  2000  ft.  distant,  by 
means  of  No.  6  B.  &  S.  wire.  If  the  energy  is  measured  at  the  power  house, 
find  the  cost  at  lOc.  per  kw.  hr.  to  light  the  lamps  8  hr.  per  day. 

Solution. — Cost  per  day  =  power  X  hr.  X  $0.10 

Power, 

P  =  PL  +  Pw 
where 

PL  —  power  required  by  lamps  =  nEI 
where 

n  =  number  of  lamps 
and 

PW  —  power  lost  in  the  wire 
Pw  =  I*R 

9 


10  ELECTRICAL  ENGINEERING 

where 

R  =  total  resistance  of  wire 

R  =  resistance  per  1000  ft.  X  1-j~ 
resistance  per  1000  ft.  of  No.  6  wire  =  0.4  ohms  at  75°F. 
Then 

R  =  0.4   X  ~     =  1.6  ohms. 


PW  =  6.62  X  1.6  =  70  watts 

PL  =  10  X  80  X  6.6  =  5280  watts 

p  =  pL  x  PW  =  5350  watts  =  5.35  kw. 
Cost  =  5.35  X  8  X  0.10  =  $4.30.     Ans. 

Such  problems  are  typical  of  existing  conditions.  An  engineer 
continually  meets  them  where  he  is  trying  to  find  what  results  are 
being  obtained  from  a  given  installation.  In  solving  them, 
accuracy  is  the  prime  consideration,  and  this  is  obtained  by  avoid- 
ing short  cuts  and  following  through,  step  by  step,  a  logical  de- 
velopment. These  problems  are  of  far  less  importance  and  inter- 
est to  the  engineering  student  than  problems  of  the  second  group. 


+  2000r  

•r  H 

\       1 

^VAAWWNAA^- 

t         J> 

1    1 

250   ^50  Kw. 

%'/////'//ti 

FIG.  6. 

As  an  illustration  of  these  take  the  following  example: 

Problem  14.  —  A  load  of  50  kw.  at  250  volts  is  to  be  supplied  by  a  power 
house  distant  2000  ft.  from  the  load.  If  the  line  costs  20c.  per  Ib.  of  copper 
laid,  find  and  plot  (a)  efficiency  of  transmission  against  size  of  wire;  (6)  cost 
of  copper  against  size  of  wire;  (c)  efficiency  of  transmission  against  cost  of 
copper. 

load  50,000 

load  +  line  loss  =  50,000  +  /»* 


50,000 
.  .  Efficiency  =  50)000 


X  resistance  per  1000  ft.  =  4  X  r 
X  wt.  per  1000  ft.  =  4w. 


FORM  OF  WORK 


11 


Tabulation : 


Wire  No.  (B.  &  S.) 

0000 

00 

1 

4 

8 

12 

16 

r  per  1000ft.1  
R  =  4r 

0.049 
0  196 

0.078 
0  312 

0.125 
0.50 

0.25 
1.0 

0.64 
2.56 

1.60 
6.4 

4.0 
16.0 

Wt.  per  1000  ft  =  w 

wt.  =  4»....  '...;. 

Cost  at  $0.20  

641 
2,564 
512.8 

403 
1,612 
322  .4 

253 
1,012 
202.4 

126 
504 
100.8 

50 
200 
40 

20 
80 
16 

7.9 
31.6 
6.32 

40  00072  =  72# 

7,840 

12,480 

20,000 

40,000 

102,400 

256,000 

640,000 

50,000  +  40,000/2  .  . 
Efficiency 

57,840 
0.865 

62,480 
0.8 

70,000 
0.715 

90,000 
0.55 

152,400 
0.33 

306,000 
0.16 

690,000 
0.072 

The  curves  are  plotted  in  Fig.  7. 


600    100  1 
90 
400      80 
70 

300  |60 
~   1 
i   IB  50 
J.I 

200^40 

30 
100      20 
10 

0        0 
OC 

< 

v 

^ 

.-C09] 

1  —  — 

^.    — 

\ 

\ 

^ 

-^ 

>< 

X 

V 

7 

\ 

\ 

A 

I 

\ 

%! 

\ 

/ 

\ 

^. 

\ 

/ 

N 

\ 

X 

\ 

[ 

^ 

"^^. 

•—  — 

*" 

^ 

•  1    > 

00    00        1        3        5        7        9       .11       13       15       17 
Size  of  Wire 
)                100               200              300             400               500 
Cost.  $ 

FIG.  7. 

Summary. — The  curves  show  (1)  efficiency  of  transmission 
decreases  as  wire  becomes  smaller,  at  first  slowly,  then  rapidly, 
and  then,  for  very  small  wires,  slowly  again;  (2)  the  cost  of  wire 
decreases  as  wire  becomes  smaller,  at  first  very  rapidly,  then  more 
and  more  slowly;  (3)  efficiency  increases  with  cost,  rapidly  at 
first,  for  low  efficiencies  and  costs,  then  more  and  more  slowly. 

1  From  wire  tables. 


12  ELECTRICAL  ENGINEERING 

It  is  evident  that  this  problem  could  be  greatly  extended  so  as 
to  include  other  variables,  such  as  current  density  in  the  wire, 
cost  of  lost  energy,  etc.,  and  indeed  it  is  characteristic  of  this 
type  of  problem  that  there  are  always  suggestive  lines  of  investi- 
gation which  tend  to  stimulate  the  student's  interest. 

The  work  of  solving  the  problem  may  be  divided  into  a  number 
of  parts,  thus:  (1)  statement  of  problem,  (2)  diagram  of  circuit, 
(3)  analytical  work,  (4)  tabulation  of  values,  (5)  plotting  of 
curves,  (6)  summary,  or  statement  in  words,  of  the  results 
obtained. 

The  statement  of  the  problem  should  be  concise.  The  dia- 
gram should  be  an  illustration  of  the  statement,  and  should  con- 
tain the  symbols  to  be  used. 

The  analytical  work  should  be  carried  out  as  far  as  possible 
with  symbols  before  the  numerical  values  are  substituted.  In 
the  above  example  there  is  very  little  opportunity  for  the  use  of 
symbols,  owing  to  the  shortness  of  the  problem.  In  later  prob- 
lems this  feature  will  be  more  apparent. 

Tabulation  should  be  arranged  with  care,  and  should  be  planned 
so  that  columns  can  be  conveniently  added.  As  a  rule,  it  is  well 
to  assign  along  the  horizontal  various  values  of  the  independent 
variable,  and  proceed,  step  by  step,  to  the  dependent  variable. 
As  in  this  case,  there  may  be  different  combinations  of  variables, 
as  number  of  wire,  cost,  and  efficiency.  This  makes  the  tabula- 
tion more  complex,  as  it  would  be  by  any  other  procedure,  but  it 
is  still  entirely  clear.  The  plotting  of  curves  is  then  carried  out, 
and  this  should  be  done  neatly  and  preferably  in  ink. 

The  problem  should  then  be  completed  with  a  brief  statement 
of  the  results  obtained.  It  is  not  always  easy  to  make  students 
take  this  last  step,  but  they  should  be  required  to  do  so,  and  to 
follow  this  general  plan  throughout,  until  they  have  formed  the 
habit  of  doing  it  and  need  no  further  compulsion. 

There  may  be  other  ways  of  working  these  problems  efficiently, 
but  it  seems  justifiable  to  urge  teachers  and  students  to  adopt  this 
method  in  preference  to  any  other  to  which  they  are  accustomed. 
It  will  insure  uniformity  and  logical  arrangement,  will  make  cor- 
recting easy,  and  will  commend  itself  to  the  student  as  well  as 
the  teacher. 

In  working  problems  of  this  nature  there  are  other  objects  than 
merely  to  illustrate  and  enforce  the  theory. 

Great  stress  is  laid  on  them,  not  only  for  the  engineering  knowl- 


FORM  OF  WORK  13 

edge  which  they  contain,  but  because  of  their  structure,  which,  it 
is  believed,  strongly  tends  to  develop  those  qualities  most  essen- 
tial in  an  engineer.  For  instance,  the  mathematical  develop- 
ment calls  for  insight  and  understanding,  the  tabulation  calls 
for  concentration  of  mind,  the  summation  of  results  calls  for 
accuracy,  and  a  study  of  the  plotted  curves  calls  for  judgment. 
At  the  same  time,  efficiency,  the  keynote  of  the  engineer,  would 
be  lacking  if  the  problems  were  not  done  in  the  shortest  and  best 
way  consistent  with  obtaining  the  desired  results,  and  it  is  obvious 
that  many  hours  will  be  wasted,  both  to  student  and  instructor, 
unless  the  work  is  done  with  order,  accuracy  and  neatness. 


CHAPTER  III 
MAGNETISM 

FARADAY  explained  magnetic  phenomena  by  assuming  that 
surrounding  a  magnet  or  a  wire  carrying  current  were  lines  of 
force. 

The  stronger  the  magnet  or  current,  the  stronger  is  the  magnetic 
field,  that  is,  the  more  lines  of  force  per  square  centimeter. 

The  introduction,  then,  of  a  magnet  into  a  space  means  the 
establishing  of  a  field  of  force. 

To  get  quantitative  ideas  about  field  strength  he  made  use  of 
the  symbol  H  which  was  called  the  intensity  of  the  field,  or  the 
force  on  unit  pole  placed  in  the  field.  It  seems  an  unfortunate 
term  since  intensity  and  density  are  readily  confused. 

B,  the  density  of  the  field,  or  the  number  of  lines  of  force  per 
square  centimeter,  is  proportional  to  H,  and  also  to  a  quantity  n, 
the  permeability  or  magnetic  conductivity  of  the  medium  in 
which  the  intensity,  H,  exists.  Thus 

B  =  fj.H. 

=  1,  it  follows  that  the  number  of  lines  of 
force  per  square  centimeter  is 
numerically  the  same  as  the 
intensity  of  the  magnetic 
field  H. 

H,  the  force  per  unit  pole, 
is  expressed  in  dynes. 

The  force  exerted  on  a  pole 
„  not  of  unit  strength,  but  of 

Jr  IG.   o.  . 

strength  m,  is 
F  =  mH  dynes, 

where,  of  course,  H  is  caused  by  other  poles  than  m. 

Consider,  now,  an  isolated  elementary  pole  of  strength  m,  from 
which  n  lines  of  force,  per  unit  pole,  protrude  radially  and  uni- 
formly in  all  directions  (Fig.  8). 

14 


MAGNETISM 


15 


At  a  distance  r  from  ra,  no  matter  what  the  medium  is  provided 
it  is  uniform,  the  density  of  the  field  is  B  —  T  —  %,  since  the  area 

of  a  sphere  of  radius  r  is  47rr2.    The  force,  H,  on  unit  pole  is  then 

_,        ,     „  ,          .    jr, 

Thus  the  force  on  pole  mi  is  r 


B 


nm 


COULOMB,  working  in  air,  found  experimentally  that  the  force 
between  two  poles  of  strength  m  and  mi  could  be  expressed  by 


p  =  k  — y-,  and  he  would  have  found  F  =  k  —^  had  he  experi- 
mented in  a  medium  of  permeability,  /*. 

Therefore,  k  may  be  written  unity  if  n  =  4ir.  In  other  words, 
if  it  is  assumed,  as  is  the  case,  that  4?r  lines  protrude  from  unit  pole. 

GAUSS  came  to  the  same  conclusion  from  another  point  of  view, 
and  the  relation  <j>  =  4irm  is  called  GAUSS'S  theorem. 

In  words,  GAUSS'S  theorem  states  that  from  a  pole  of  strength 
m  radiate  outward  47rm  lines  of  force,  or  the  total  outward  flux, 
$,  from  pole  m  is  4irm  lines.  1 

Cylindrical  Poles. — To  find  the  intensity  of  the  magnetic  field 
H  at  a  point  distant  r  from  a  uniform  cylindrical  pole  of  strength 
m  (Fig.  9).  By  GAUSS'S  theorem  the, flux  <f>  =  47rm,  and  #  = 

The  area  of  a  cylinder  of  radius  r  and  length  I  is 

,       .    _  2m 

at  p,  is 


_    flux 
area* 

2irrl.     Thus 


2m        , 
=  ->  and 


\ 

1 

t 

I 

FIG.  9. 


FIG.  10. 


Flat  Poles. — To  find  the  intensity  of  the  field,  H ,  at  a  point  dis- 
tant d  from  one  side  of  a  flat  pole  of  strength  m  (Fig.  10).  As- 
sume that  the  lines  of  force  are  perpendicular  to  the  surface.  Let 
B  =  flux  density  at  any  distance.  Then  the  flux  coming  from 

one  of  the  surfaces  of  the  magnet  is  <£  =  — ^—  =  2irm. 

4 

If  the  area  of  the  pole  face  is  S.  then  B  =  — «->  and  H  =  — FT- 

S  MO 

1  For  a  more  complete  discussion  of  GAUSS'S  theorem  see  "Advanced  Course 
in  Electrical  Engineering." 


16  ELECTRICAL  ENGINEERING 

Magnets  as  Commonly  Used  in  Meters.  —  To  find  the  magnetic 
intensity  between  poles  (Fig.  11).  Let  S  be  the  area  of  a  pole 
face  and  d  the  distance  between  poles. 

The  density  in  the  gap  between  the  two  pole  faces  is  due  to  the 
magnetic  north  pole,  N,  as  well  as  to  the  south 
pole,  S. 

If  the  lines  of  force  flow  outward  from  the 
north  pole,  they  flow  inward  from  the  south 
pole.  Thus  a  simple  examination  will  show 
that  the  fluxes  add  in  the  gap  and  cancel  each 
other  in  the  outside  region. 

The  total  flux  from  N  is  4irm  and  one-half  of 
this  flux  is  assumed  to  be  in  the  gap,  the  other  half  extending 
outward. 

The  density  in  the  gap  due  to  N  will  then  be 

_  4?rm  _  2-rrm 
^n  ":  ~2S~      ~S~' 
Similarly,  due  to  S, 


and  „  =  #  =  4?rm 

M         Bfj. 

In  all  practical  problems  where  magnets  act  in  air  only,  /*  is, 
of  course,  unity. 

Consider,  now,  the  pull  between  the  faces  of  a  magnet  as  shown 
in  Fig.  11. 

The  flux  density  at  the  south  pole  due  to  the  flux  from  the  north 

.    .     n         2irm  2irm      ,  .  , 

pole  is  Bn  =  —g—    .  .  H  =  —  -g-  =  force  on  unit  pole  at  the  sur- 

face of  the  south  pole. 

Since  the  south  pole  has  a  strength  m,  the  force  on  it  is 
therefore 

'  do) 


Usually  the  density,  B,  in  the  gap  is  known. 

Substituting  the  value  of  m  from  (9)  into  (10)  gives 
27r 

~ 


4V 
or  the  force  in  dynes  per  sq.  cm.  is  F0  =  ^  —  • 


MAGNETISM  17 

In  air,  where  ju  =  1, 

B2 

F0  =  £-  dynes  per  sq.  cm.  (11) 

In  Ib.  per  sq.  in.,  the  formula  becomes 


It  is  seen  that  if  the  pole  strength,  m,  remains  the  same  while  the 
faces  of  the  magnet  approach  each  other,  the  density,  and  thus 
the  force,  is  constant. 

The  work  done  is  then  Fd,  where  d  is  the  distance  between  the 

TT/       B*Sd 
poles,  or  W  =     ~     • 

Energy  Density  in  a  Field.  —  The  volume  of  space  through 
which  the  body  is  moved  is  Sd.  The  energy  density,  or  joules 
per  cu.  cm.  of  space  between  poles,  is  then  : 

1        B2  Bz 

*  §3  =  S  ergs  =  8000*  J0ules' 

The  conception  of  energy  density  is  merely  mentioned  at  this 
point.  Similarity  of  magnetic  and  eledtric  fields  will  be  shown 
later  on  together  with  the  development  of  theory  and  problems 
in  electro-statics. 

Limits  of  Pole  Intensity.  —  In  practice  it  is  found  that  the  limits 

to  which  pole  intensity  -~-  can  be  pushed  are  as  given  in  the  fol- 
lowing table: 

TABLE  II.  —  APPROXIMATE  LIMITING  VALUES  OF  •„• 

o 

For  wrought  iron  magnets,       1600  units  of  pole  strength  per  cm. 
For  soft  steel  magnets,  1600  units  of  pole  strength  per  cm. 

For  cobalt  magnets,  1300  units  of  pole  strength  per  cm. 

For  nickel  magnets,  500  units  of  pole  strength  per  cm. 

For  permanent  steel  magnets,    800  units  of  pole  strength  per  cm. 

The  Magnetic  Cycle.  —  According  to  the  molecular  theory  of 
magnetism,  magnetic  bodies  are  composed  of  minute  magnets 
which  attract  and  repel  each  other,  and  which  are  partly  free 
to  turn  under  the  influence  of  magnetizing  forces.  When  strongly 
magnetized,  these  molecular  magnets  are  pointed  in  the  direction 
of  the  magnetic  force.  When  the  force  is  removed,  they  still 
tend  to  point  in  the  same  direction,  and  thus  the  body  exhibits 
magnetization,  which  is  called  residual  magnetism. 


18  ELECTRICAL  ENGINEERING 

The  magnetic  state  of  a  body  is  shown  with  reference  to  the 
"magnetizing  force"  by  a  curve  called  the  hysteresis  loop  (Fig. 
12). 

Magnetization  of  an  iron  bar  is  ordinarily  accomplished  by 
sending  current  through  a  number  of  turns  of  wire  wound  around 
the  bar.  The  magnetization  is  thus  produced  by  the  ampere- 
turns  (A.T.).  The  number  of  lines  of  flux  set  up  per  unit  area 
enclosed  by  the  turns  will  with  a  long  bar  be  shown  to  be 

r  /    'M  =  B,  where  ju  is  the  permeability  of  the  bar  and  I  is 

its  length.     Since  in  air  /-t  =  1  and  H  =  B,  it  follows  that  the 
intensity  of  the  magnetic  field  in  a  solenoid  is: 

QAirA.T. 


The  hysteresis  loop  is  drawn  with  flux  density,  B  (in  lines  per 
square  centimeter  or  per  square  inch),  as  ordinates  and  the  mag- 
netic field  intensity,  H  (or  frequently,  for  convenience,  ampere- 

TT7\ 

turns  per  inch  length  of  magnetic  circuit,  j—  J  ,  as  abscissae. 

The  construction  of  the  loop  is  as  follows:  Imagine  a  bar  of 
iron  wound  with  many  turns  of  insulated  wire.  If  the  iron  has 
no  residual  magnetism  at  the  beginning,  be- 
fore current  is  sent  through  the  wire,  there 
will  be  no  magnetizing  force  and  no  flux,  and 
consequently  the  first  or  starting  point  on  the 
curve  will  be  at  a  (Fig.  12).  As  more  and 
more  current  is  sent  through  the  wire,  that 
is,  as  the  magnetizing  force  is  increased  pro- 
portionally to  the  current,  the  flux  or  induc- 
tion  density,  B,  is  increased,  not  according 
to  a  simple  law,  but  in  such  a  way  as  to  give 
the  characteristic  curve  (1)  from  a  to  6. 

If  the  magnetomotive  force  (m.m.f.)  expressed  in  ampere- 
turns  is  now  decreased,  the  curve  (1)  is  not  retraced,  but  B 
follows  curve  (2)  from  b  to  c.  At  c,  H  =  0,  while  B  continues 
to  have  a  value  represented  by  the  line  ac.  This  value  of  B 
corresponds  to  the  residual  magnetism  of  the  iron. 

If,  now,  the  current  be  reversed,  so  that  H  is  given  negative 
values,  B  continues  to  decrease  from  c  to  d.  At  the  point  d, 
B  =  0,  while  H  has  the  negative  value  ad.  This  value  of  H  is 


MAGNETISM  19 

called  the  "coercive  force"  of  the  magnet.  It  is  the  magnetizing 
force  necessary  to  reduce  the  remanent  magnetism,  ac,  to  zero. 
As  H  is  further  increased,  negatively,  B  follows  the  curve  de. 
At  6,  which  corresponds  to  b  with  positive  H,  the  current  is  again 
reduced,  and  B  follows  curve  (3)  to  /,  which  gives  the  value,  of, 
of  negative  remanent  magnetism  corresponding  to  ac  for  H  =  0. 

Thus,  the  point  a  is  not  reached  again,  but  as  H  is  now  given 
increasing  positive  values,  the  curve  goes  through  g  to  b,  complet- 
ing the  loop. 

In  obtaining  a  single  loop,  the  points  do  not  usually  come  into 
such  close  agreement,  due  primarily  to  the  fact  that  there  is 
always  some  remanent  magnetism  at  starting,  which  prevents 
the  curve  from  beginning  exactly  at  a.  But  in  the  case  of  many 
uniform  reversals  of  H,  as  occurs  in  electrical  machinery,  the 
loop  is  retraced  uniformly  so  long  as  the  limiting  values  of  H  re- 
main constant. 

It  will  be  later  shown,  in  connection  with  the  study  of  hysteresis 
losses,  that  the  area  enclosed  by  the  loop  is  proportional  to  the 
work  done  on  the  magnet  per  cycle. 

T> 

Permeability. — The  ratio  -p  is  called  the  permeability,  and  is 

a  measure  of  ease  with  which  lines  of  flux  are  set  up  in  a  given 
material.  Permeability  is  denoted  by  the  symbol  /*.  Numer- 
ically, B  =  H  in  air  (or  vacuum)  since  /*  =  1.  In  the  magnetic 
metals,  particularly  iron,  steel,  nickel  and  cobalt,  ju  undergoes 
wide  variation  in  value,  with  different  values  of  H . 

For  a  more  complete  discussion  of  the  subject  of  magnetism  the  student  is 
referred  particularly  to  EWINQ'S  "Magnetic  Induction  in  Iron  and  Other 
Metals." 


CHAPTER  IV 


N 


PRINCIPLE  OF  THE  ELECTRIC  MOTOR 

A  wire  carrying  a  current  was  discovered  by  OERSTED  to  be 
surrounded  by  a  magnetic  field,  which  is  strongest  near  the  wire. 
A  small  needle,  placed  in  the  field  (Fig.  13),  is  directed  along  the 
lines  of  force,  but  there  is  practically  no  tendency  for  it  to  move 
toward  the  wire  as  the  forces  of  attraction  exerted  on  its  poles 

are  equal  and  opposite.  A  long  needle, 
however,  tends  to  move  toward  the  wire 
as  there  is  a  component  of  force  on  each 
pole  in  the  direction  of  the  wire. 

A  wire  carrying  current,  placed  in  a 
field  perpendicular  to  the  lines  of  force 
(Fig.  14),  causes  the  flux  to  be  distorted, 
and  this  tends  to  force  the  wire  in  such 
a  direction  that  the  lines  shall  again 
take  up  their  normal  position.  This  is 
the  principle  of  the  electric  motor. 
The  electric  motor  consists  (Fig.  15)  of  a  number  of  wires 
wound  on  a  drum,  and  so  placed  in  a  magnetic  field  that  the 
current  is  caused  to  flow  downward  (toward  the  plane  of  the 
paper)  on,  say,  all  the  wires  adjacent  to  the  north  pole,1  and  up- 
ward on  all  the  wires  adjacent  to  the  south  pole.  The  wires  on 


FIG.  13. 


N 


N 


FIG.  14. 


FIG.  15. 


the  left,  then,  tend  to  move  downward,  and  those  on  the  right 
upward,  and  thus  rotation  is  produced. 

1  In  the  diagram  a  cross,  <8>,  is  used  to  represent  down-flowing  current  and 
a  dot,  O,  up-flowing  current  in  accordance  with  notation  in  common  use. 

20 


PRINCIPLE  OF  THE  ELECTRIC  MOTOR 


21 


The  current  which,  when  flowing  in  a  wire  1  cm.  long  placed 
at  right  angles  to  a  field  having  a  density  of  1  line  per  sq.  cm., 
gives  a  force  of  1  dyne  is  called  the  abampere. 

The  force,  in  dynes,  is  then 

F  =  IIB 

where  /  is  the  current  in  abamperes,  I  the  length  of  wire  in  centi- 
meters, and  B  the  flux  density  of  the  field  in  lines  per  square  centi- 
meter. The  force  is  due  to  the  interaction 
of  flux  and  current. 

If,  however,  the  lines  are  not  at  right 
angles  to  the  wire,  B  must  be  replaced  by 
its  component  which  is  at  right  angles  to 
the  wire.  If  the  angle  is  a  (Fig.  16),  then 
the  force  is  F  =  IIB  sin  a,  where  B  sin  a  is  the  component  of  flux 
at  right  angles  to  the  wire. 

Problem  16.  —  A  copper  wire  carrying  10  amp.  is  placed  in  a  magnetic  field 
of  10,000  lines  per  sq.  cm. 

What  is  the  force  in  pounds  on  each  centimeter  of  the  wire  (a)  if  it  lies 
perpendicular  to  the  direction  of  the  magnetic  field,  (6)  if  it  lies  parallel  to 
the  field,  (c)  if  it  makes  an  angle,  a,  with  the  direction  of  the  field? 


N 


FIG.  16. 


(a) 


Solution.  —  F  =  IIB  sin  a 

F,  per  cm.  =  IB  sin  a. 

Sin  a  =  sin  90°  =   1 

/  =  10  amp.  =  lab  amp. 
B  =  10,000 
.*.  F,  per  cm.  =  10,000  dynes. 

10,000  dynes  =  ==  10.2  grams 


10 


0.02245  Ib. 


453.6 

(6)  Sin  a  =  sin  0°  =  0 

.'.  F,  per  cm.  =  0. 

(c)    For  any  angle,  a, 

F,  per  cm.  =  10,000  sin  a  dynes 
=  0.02245  sin  a  Ib. 


Determinations  of  Magnetic  Intensity.  — 

Magnetic  intensity  at  the  center  of  a  coil 
(annulus)  .  Let  a  magnet  pole,  m,  be  placed 
at  the  center  of  a  coil  (Fig.  17).'  It  will 
send  out  lines  in  all  directions,  some  of 
which  will  strike  an  element  of  the  coil,  dl, 


22 


ELECTRICAL  ENGINEERING 


at  right  angles.     Thus  a  force,  dF,  will  be  generated  in  the  direc- 
tion of  the  axis,  as  indicated,  and  its  value  will  be 

dF  =  IBdl  =  /  ™  dl 


snce 


The  total  force  on  the  coil  will  be 


-ndl 


2irrlm 


This  will  be  the  force,  due  to  m,  with  which  the  coil  will  tend  to 
move  along  its  own  axis.  It  is  obviously  also  the  force  on  m  due 
to  the  coil.  Thus  if  a  unit  pole  (m  =  1)  replaces  the  pole  of 
strength  m  the  magnetic  field  intensity  at  the  center  of  the  coil 
is  found.  It  is: 

a_*a*,*L 

r2  r 

Magnetic  Intensity  at  Any  Point  along 
of  a  Coil.— The  force  dF  will  act 
angles  to  the  line  joining  m  and 


h— 

m 

~ 

* 
F 

ji---" 
I* 

10.  18. 

dF  =  IBdl  =  Idl 

d 

dF  has   components,  dF  cos  a  and  dF  sin  a  where  sin  a  =  — 
and 


a  — 


sin 


The  component  of  force  which  tends  to  move  the  coil  in  the 
direction  of  its  axis  is  dF  sin  «.     Call  this  component  dFi. 
Then 


and 


For 


Jp 


m 


. 
/  I.  sin3  .adZ  = 


27rm  I  sin3 


m  =  i  pl  =  H 

sin3a 


PRINCIPLE  OF  THE  ELECTRIC  MOTOR 


23 


since  the  component  dF  cos  a  is  balanced  around  the  coil  and 
thus  exerts  no  force. 

Magnetic  Intensity  in  the  Center  of  a  Long  Coil. — The  force 
at  m,   due  to  an  element  of  the  coil,  dx  (Fig.  19),  is  dF  = 


2irml  sin3  a 


,  where  /  is  the  current  in  abamperes,  in  the  ele- 


ment dx. 


FIG.  19. 

If  the  current  per  centimeter  length  of  the  coil  is  /c,  then 

,   ,._       2irmlcdx  sin3  a 
I  =  Icdx,  and  dF  = 

x                                                            /          1     \ 
But  —  =  cot  a.     Differentiating,  dx  =  r  ( =-^ — )  da. 

Substituting  this  value  of  dx  in  the  formula, 

2irmlcr  sin3  a  , 

dF  = ^ da 

r  sm2  a 

=  —  2irm7c  sin  ada. 


Co.  -  ai 

.*.  F  =    I  27rm/c  sin  ada  =  4irl, 

./a  ="  TT  —  ai 


m  cos 


or  H  =  47r/c  cos  ai. 

For  very  long  coils,  cos  ai  =  1,  and 

H    =   47T/C. 

The  relation  between  H  and  the  ampere-turns  of  a  coil  may  be 
found  as  follows: 

Let  there  be  a  current  of  I  abamp.  in  the  coil,  and  let  n  = 
number  of  turns.  Then  nl  =  abamp.-turns.  Abamp.-turns 

nl 

per  cm.  =  -r-  =  Jc,  where  I  =  length  of  coil  in  centimeters. 

47m/1 
I 


Then  H  =  4ir7c 


When    the    current    is  in  amperes, 


1  NOTE. — It  should  be  noted  that  the  above  value  of  H 


holds  only 


for  infinitely  long  solenoids  since  it  was  derived  on  that  assumption.     For 
practical  purposes,  according  to  the  accuracy  required,  this  value  of  H  may  be 


24 


ELECTRICAL  ENGINEERING 


H 


—j — ,  whence,  amp.-turns  =  0. 


SHI 


(12) 


i 

If  I  is  in  inches,  amp.-turns  =  Q.313HL 
When  the  coil  has  an  air  core,  H  is  numerically  equal  to  B,  and 
amp.-turns  =  Q.SBl  or  =  0.313J5Z.1 


FIG.  20. 


FIG.  21. 


Application  of  Magnetic  Formulae  to  Instruments. — Let  a  rec- 
tangular coil  of  height,  a,  and  width,  6,  be  suspended  in  a  magnetic 
field  of  uniform  density,  B  (Fig.  20).  The  two  sides,  a,  are  per- 
pendicular to  the  flux,  and  therefore,  with  a  current  of  /  abamp., 
there  will  be  a  force  on  each  wire  of  Fa  =  IBl  =  IB  a  dynes. 


FIG.  22. 

This  force  will  produce  a  torque  around  the  axis,  on  each  wire,  of 
T0  =  IBa  ~  dyne-cm.  The  total  torque  per  turn  =  2Ta  =  I  Bab, 

£i 

and  if  there  are  n  turns,  T  =  InBdb,  dyne-cm. 

In  practical  instruments,  T  should  be  about  1  gram-cm. 

Let  a  circular  coil  of  radius  r,  as  in  Fig.  21,  be  suspended  in  the 
field.  To  find  the  torque  on  any  element  dl.  The  useful  part 

used  whenever  it  is  desired  to  find  the  magnetic  field  intensity  along  the  axis 
of  a  solenoid,  and  not  very  near  the  ends,  provided  the  length  of  the  coil  is 
about  50  times  its  diameter. 

It  is  necessary  to  observe  this  (always  depending  on  the  accuracy  desired) 
on  account  of  the  disturbing  effects  of  the  ends,  as  can  be  easily  seen  by 
comparing  the  figures.  (Fig.  22.) 

1  NOTE. — It  can  be  proven  that  the  density  in  the  middle  of  such  long  coil 
is  uniform,  thus  if  A  is  the  area  inside  of  the  solenoid  the  total  flux  is  AB, 
area 


PRINCIPLE  OF  THE  ELECTRIC  MOTOR  25 

of  dl  is  its  component  perpendicular  to  the  lines  of  flux,  =  dl  sin  6. 
Then,  dF  =  InB  sin  Bdl.  This  force  acts  with  a  lever  arm  =  r 
sin  6,  and  the  torque  is  therefore  dT  =  InB  sin  6  X  r  sin  0dl.  But 


dl  =  rd0.     Hence  dT  =  InBr*  sin2  0d0,  and  T  =    I  7n£r2  sin2 

Jo 

[01  T27r 

-£  -  7  sin  20 
Zl        4  J  0 

=  InBr*ir  =  InB Ay  where  A  =  area  of  the  loop. 

In  practice,  permanent  magnets  are  generally  used  to  produce 
the  flux. 


CHAPTER  V 
DESIGN  OF  A  LIFTING  MAGNET 

It  has  been  shown  that  for  a  path  of  magnetic  lines  in  air, 
the  following  relation  obtains:  amp.-turns  =  0.313  Bl",  if  inch 
measurements  are  used.  For  an  iron  path,  the  necessary  ampere- 
turns  are  obtained  from  a  curve  of  B  vs.  AT,  where  B  is  the 
flux  density.  Such  a  curve  called  either  magnetization  or  "  satu- 
ration "  curve,  is  obtained  experimentally  from  a  sample  of  any 
desired  magnetic  material.  The  curve  thus  obtained  will  be 
approximately  correct  for  that  material,  but  variations  are  always 


100  120 

Ampere-Turns  per  Inch 

FIG.  23. 


juo 


160 


180 


liable  to  occur  due  to  either  physical  or  chemical  influences  by 
which  any  portion  of  the  material  is  made  to  differ  from  the 
sample  used  to  derive  the  curve. 

By  testing  many  samples  a  typical  curve  is  obtained  for 
any  given  material.  In  Fig.  23  is  given  a  set  of  these  satura- 
tion curves  of  iron  and  steel  as  commonly  employed  in  electrical 
machinery. 

26 


DESIGN  OF  A  LIFTING  MAGNET 


27 


Let  it  now  be  required  to  make  the  calculations  for  the  design 
of  a  cast-iron  electromagnet  to  lift  a  weight  of  1000  Ib.  through  a 
gap  of  1.5  in.  Let  it  be  assumed 
that  the  magnet  core  is  of  the 
shape  and  dimensions  given  in 
Fig.  24.  Then  the  area  of  a  pole 
face  is  10  X'  5  =  50  sq.  in.  The  I 
two  pole  faces  have  an  area  of 
2  X  50  =  100  sq.  in.  Then  the 
weight  to  be  lifted  per  square  inch 
of  area  is 


<M 


1000 Ibs, 

FIG.  24. 


Wt.  per  sq.  in. 


1000 
100 


=  10  Ib.  =  F. 


But,  from  (11),  F  = 


72,134,000 


.*.  B,  the  flux  density  in  air,  =  V721,340,000  =  26,800  lines 
per  sq.  in.     From  (12),  the  amp.-turns  required  for  each  gap  = 
0.313  HI  in.  =  0.313  X  26,800  X  1.5  =  12,600. 
AT  required  for  two  gaps  =  2  X  12,600  =  25,200. 
AT  required  for  the  iron,  from  the  magnetization  curve  for 
cast  iron  =  32  per  in.  length  of  the  flux  path  in  the  iron. 
Length  of  mean  path  in  the  iron  =  70  in. 
.'.  Total  AT  required  for  the  iron  =  70  X  32  =  2240,  and  total 
AT  for  both  air  and  iron  =  25,200  -f  2240  =  27,440. 

It  is  now  necessary  to  arrange  the  winding  so  that  the  heat 
developed  by  the  current  in  the  coil  shall  not  cause  an  excessive 
temperature  in  the  coil.  " 

Assuming  a  permissible  power  loss  of  0.4  watt  per  sq.  in.  of 

exposed  coil  surface,  an  estimate  can 
be  made  of  the  proper  amount  of 
space  to  be  occupied  by  the  coil. 
Assuming,  as  a  guess,  a  depth  of  wind- 
ing of  3  in.  and  a  coil  length  of  20  in. 
(Fig.  25),  the  exposed  surface  of  the 
coil  is  1270  sq.  in.  The  total  watts  developed  should  then  be 
0.4  X  1270  =  508.  Assuming,  further,  that  the  magnet  is  to  be 
designed  for  operation  on  a  100-volt  circuit,  the  current  is 
508 

loo  =  5-08  amp- 


Cross-section  of  Coil 


28  ELECTRICAL  ENGINEERING 

27  440 
The  number  of  turns  =  -^  -  =  5400,  and  the  resistance  of 

O.Uo 

.,       W         533          ,.,_    , 
the  coil  =  yj  =  ,-  oo\2  =  19.7  °nms. 

The  mean  length  of  one  turn  may  be  estimated  to  be  42  in. 

42 
Then,  total  length  of  wire  =  5400  X  TS  =  18,900  ft. 

i  si  , 

19  7 
Res.  per  1000  ft.  =  =  1.04  ohms. 


From  tables,  the  nearest  size  wire  is  No.  10.  Diameter  of 
No.  10,  with  double  cotton  covering,  is  0.112  in.  It  must  then 
be  found  out  if  there  will  be  room  enough  for  the  turns  in  the 
space  allowed. 

Problem  16.  —  Assume  cast  iron  to  cost  l^c.  per  Ib.  and  copper  16c.  per 
Ib.  Find  the  least  cost  of  a  magnet,  of  any  desired  shape,  to  lift  1000  Ib. 
with  a  gap  of  1.5  in. 

Use  0.4  watt  per  sq.  in.  of  coil  surface  as  permissible  power  loss. 


A 


dx 


CHAPTER  VI 

GENERATION  OF  ELECTROMOTIVE  FORCE  IN  A 
DYNAMO 

FARADAY  found  in  1831  that  the  electromotive  force  produced 
in  a  circuit  was  proportional  to  the  rate  of  change  of  the  lines 

of  force  enclosed  in  the  circuit.     That  is,  he  found  that  e  —  k  -57* 

This  discovery  was  really  the  foundation  upon  which  electrical 
engineering  was  built. 

The  truth  of  the  relation  may  be  seen  by  considering  a  rec- 
tangular loop  of  wire  carrying  current  I 
moved  a  short  distance  dx  in  time  dt,  the 
motion  being  at  right  angles  to  the  direction 
of  the  lines  of  force  (Fig.  26).  1 

The  force  on  1  cm.  of  wire,  A,  which  is  in 
the  field  is  F  =  IB.  Thus  the  mechanical 
work  done  in  moving  the  loop  from  AB  to 
A'B'  is  Fdx  =  IBdx.  F7G,  26< 

As  has  been  shown,  the  electrical  work  is 

eldt  and  the  mechanical  and  electrical  work  must  be  equal  and 
opposite. 

dx 
.'.IBdx  =    —  eldt  or  e  =   —  B  -57-     But  Bdx  is  the  change  of 

flux  d<j>  enclosed  in  the  loop.     Thus 

j      ," -  "'•''.-.  e=~dit  ' 

Consider  now  that  the  loop  revolves  in  a  uniform  magnetic 
field.  When  the  loop  encloses  the  entire  field  it  may  be  said 
to  be  in  the  zero  position.  Let  it  be  assumed  that  in  zero  posi- 
tion it  encloses  100  lines.  In  position  1,  displaced  10°,  it  will 
then  enclose  98.5  lines.  The  loss  of  1.5  lines  from  the  loop  has 
resulted  in  a  generated  e.m.f.  or,  in  general, 

dt  t%  —  ti  At 

29 


30 


ELECTRICAL  ENGINEERING 


If  the  coil  rotates  at  the  rate  of  1  r.p.s.,  the  time  required  for  it 
to  move  10°  is       :  of  1  sec.  =  0.0277  sec.  =  A£. 


Then 


e  =  — 


1*5 


0.0277     "  0.0277 


=  54a6  volts. 


This  procedure  may  be  followed  and  a  tabulation  made  for 
every  10°,  so  as  to  obtain  data  from  which  to  plot,  point  by  point, 
an  e.m.f.  wave,  thus: 


Angular  position  of  coil 

0° 

10° 

20° 

30°,  etc. 

Flux  enclosed,  <£  

100 

98.5 

94.0 

86  6 

Change  of  flux,  <f>z  —  <£i                             ... 

—  1.5 

—4  5 

—7  4 

A0 

a    —     

54.0 

163.0 

267  0 

A« 
Plotted  at  angle  

5° 

15° 

25° 

Values  of  e.m.f.  are  plotted  at  angles  given  in  the  last  line,  that 
is,  at  the  mid-angular  divisions,  since  they  represent  average 
values  of  e  over  each  10°  of  displacement  of  the  loop. 

Problem- 17. — Assuming  the  field  to  be  uniform,  carry  out  the  procedure 
as  just  indicated  for  a  complete  rotation  of  the  loop,  and  show  by  plotting 
volts  against  angular  displacement  that  the  curve  is  a  sine  wave. 

E.m.f.  Waves  in  Fields  that  are  not  Uniform. — Consider  a 
field  between  rounded  poles  of  radius  r, 
distant  2r,  in  which  a  coil,  of  width  2r, 
revolves  (Fig.  27). 

The  field  will  be  most  dense  in  the 
middle.  The  density  will  be  assumed  to 
be  inversely  proportional  to  the  distance 
between  poles.  To  find  the  density  along 
any  line  a&.  The  length,  db  =  4r  —  2r  sin  6. 


Then 


B 


4r  —  2r  sin  0 


where  k  is  some  constant — i.e.,  proportionality  factor. 

This  equation  holds  only  for  6  from  o  to  TT,  in  a  revolution. 

k 

If  B  is  taken  from  TT  to  2ir.  B  =  -A — r-= — : 

4r  +  2r  sin  6 


GENERATION  OF  ELECTROMOTIVE  FORCE       31 

When  the  coil  moves  a  distance  ds,  there  is  a  change  d<{>,  in 
the  amount  of  flux  enclosed  by  the  coil,  per  unit  length  of  the 
coil  parallel  to  the  shaft.  d<fr  is  propor- 
tional to  the  component  of  ds  at  right 
angles  to  the  direction  of  the  flux,  or  to  dx 
(Fig.  28). 

Thus,  d<j>  =  2Bdx  is  the  change  in  flux  per 
centimeter  length  of  coil,  due  to  both  con- 
ductors. Then  d<f>  =  —  2Brd0  sin  0,  since 
ds  =  rd0,  and  dx  —  —  ds  sin  0. 

Substituting, 

kr  sin  0d0  k  sin  0d0 

nax  =    — 


FIG.  28. 


4r  —  2r  sin  0 


4  -  2  sin  0 


Then, 


_  d<j>  __  JL^    /2k  sin  0d0  \         k  s 
e  "  ""  dt  ~  dt    U  -  2  sin  07  "  2T^ 


fc  sin  0 


sn        £ 

Sin  0  may  be  written  sin  2irnt,  where  0  is  expressed  in  radians 
and  2irn  denotes  angular  velocity.  Then  -j=r  =  27rn,  and  n  is 
in  revolutions  per  second,  or  is  frequency  jn  a  two-pole  machine. 
For  machines  of  any  number  of  poles  -jj  =  2wf  where  /  = 
frequency. 


k  sin  0 
2  -  sin  0 


X2wf. 


Let 


Then 


and 


100  volts.     This  occurs  when  0  =  ~ 


100  = 


Substituting  this  value  of  k, 

_  emax      2-n-f  sin  0  _  emax  sin  0 
=  27r/  X  2  -  sin  0  "  2  -  sin  0 ' 

Problem  18. — Calculate  and  plot  the  e.m.f.  for  th$  above  condition,  for 
one-half  wave. 

E.m.f.  Wave  when  the  Coil  is  Wound  on  an  Iron  Core. — In 

all  these  cases  it  is  sufficiently  correct  to  consider  only  the  lengths 


32 


ELECTRICAL  ENGINEERING 


of  the  flux  path  in  the  air.  By  following  the  general  procedure  of 
the  preceding  paragraph,  the  e.m.f.  of  a  coil  on  the  iron  core  is 
found  to  be 

(m  —  1)  sin  0 


where 


m  -  sin  0 
2mr 


m  =  7^-'  or  D 


Problem  19. — Calculate  and  plot  for  one-half  wave,  the  e.m.f.  for  this 
case,  when  emax  =  100  and  m  =  1.1.  (Fig.  29.) 

Additional  Problems  for  the  Determination  of  E.m.f.  Waves. — 
It  is  very  good  experience  for  the  student  to  work  out  and  plot 
a  number  of  these  waves.  For  this  purpose  a  few  additional 
problems  are  suggested. 

Problem  20. — Determine  the  e.m.f.  of  a  coil  wound  on  a  wooden  drum 
when  Bmax  =  100  lines  per  sq.  cm.,  speed  =  1  r.p.s.  and  the  dimensions  of 
the  dynamo  are  as  given  in  Fig.  30.  Dimensions  are  given  in  centimeters. 
Plot  the  wave,  point  by  point,  for  each  millimeter  of  distance  across  the  pole 
face. 


_L 


Problem  21. — On  an  iron  armature  between  rectangular  poles  as  in  Fig.  31, 
let  two  coils,  at  right  angles  to  each  other  (that  is,  in  space  quadrature),  be 
joined  in  series,  so  that  their  e.m.f.  waves  add.  Plot  the  resultant  e.m.f.  per 
centimeter  length  of  the  armature.  Show  that  the  resultant  e.m.f.  due  to 
the  two  coils  is  less  than  their  sum. 

Problem  22. — Same  as  last,  but  for  three  coils  spaced  120°  apart. 

Problem  23. — Continue  the  development  of  the  method  of  the  last  two 
problems  and  finally  obtain  the  average  value  of  the  e.m.f.  of  an  armature 
whose  conductors  are  spaced  uniformly  around  the  periphery. 


CHAPTER  VII 


FIG.  32. 


INDUCTANCE 

Inductance.  —  When  a  circuit  connected  to  a  source  of  e.m.f.,  et 
is  closed  through  a  switch,  S  (Fig.  32),  a  current  is  established 
in  the  coil,  and  sets  up  a  magnetic  flux 
which  links  with  the  turns  of  the  coil. 
This  flux  produces  a  back,  or  counter 

e.m.f.  in  each  turn,  e  =  —  -TJ-,  or  in  the 

N  turns,  e  =  —  N  -£•    Expressed  in  volts 

this  is 

N_  d$ 

108   dt 

Inductance  is  defined  as  the  number  of  iijterlinkages  of  flux  with 
turns,  per  unit  current,  or,  in  symbols, 


Expressed  in  practical  units  it  is: 


e  =  — 


(13) 


L  = 


where  i  is  the  current  in  amperes. 

Problem  24.  —  Let  a  coil  of  200  turns  be  supplied  with  various  amounts  of 
current,  and  let  the  flux  produced,  when  1  amp.  flows,  be  1000  lines.  Find 
the  inductance.  Tabulating  from  Eq.  (14): 


i 

i 

2 

4 

10 

tf).  . 

1000 

2000 

4000 

10000 

N  
N<f>  

200 
0.2  X  10« 

200 
0.4  X  106 

200 
0.8  X  106 

200 
2  X  10fl 

T  

0.2  X  106 
0.002 

0.2  X  106 
0.002 

0.2  X  106 
0.002 

0.2  X  10« 
0.002 

1  Students  almost  invariably  have  difficulties  with  inductance.     This 
problem  is  given  solely  for  the  purpose  of  impressing  upon  them  the  fact  that 
inductance  is  a  "constant"  of  the  circuit. 
3  33 


34  ELECTRICAL  ENGINEERING 

From  the  values  of  I/,  thus  obtained,  it  is  seen  that  L  is  a 
constant,  and  is  independent  of  the  current;  it  is  a  "  circuit  con- 
stant" similar  to  resistance  in  a  coil  having  a  non-magnetic  core. 

Transposing  Eq.  (14), 


"J&- 

Differentiating  with  respect  to  time, 

di  _    N    d<j> 
LJt"l08~dt 
Substituting  into  (13), 


which  is  the  common  expression  for  induced  electromotive  force, 
or  counter  e.m.f.  of  self-induction.1 

In  Fig.   33  is  represented  a  circuit  of 
resistance,  r,  and  inductance,  L,  which  is 
oj      connected  to  a  source  of  e.m.f.,  e.     When 

1  *-  o 


*'  the  switch,   S,  is  closed,   the  e.m.f.  has 

Tfl     _         qq 

to  overcome  the  resistance,  r,  and  also 
the  counter  e.m.f.  of  self-induction  due  to  the  setting  up  of  flux 
in  the  coil.  Therefore  we  may  write: 

e  =  ir  +  L^  (15) 

This    equation    is    fundamental,    and    is    general   for  circuits 
possessing  only  resistance  and  inductance  of  constant  value. 

An  algebraic  relation  between  the  impressed  e.m.f.  and  the 
current — assuming  in  this  case  that  the  e.m.f.  is  kept  constant 
— is  found  as  follows: 


e-ir 


1  In  an  ironclad  magnetic  circuit  the  inductance  is  not  a  constant.     It 
depends  upon  the  permeability. 

The  flux  is  not  proportional  to  the  current  producing  it  but  is  a  complicated 

function  thereof.    In  that  case  —  ^  =  ~  (Li)  =  L  ~ '  +  i^. 


INDUCTANCE  35 

/.  t  =  -  ?  log  (e  -  ir)  +  C, 


whence 

rt 
log  (e  -  ir)  =   -  £  +  Ci, 

and 

"  +  Cl        „    -r^ 
e  —  ir  —  €    L          =  C2€    L 

(16) 

C2  is  determined  from  the  nature  of  the  problem. 
The  rate  of  energy  supply  or  power  equation  corresponding 
to  (15)  is  obviously 

ei  =  izr  -f  Li 


r 
\ 

Jo 


and  the  energy  supplied  by  the  generator  is 

T 

eidt. 

The  energy  dissipated  in  heat  is 

f*T 

\    i*rdt 

Jo 

and  the  energy  supplied  and  thus  stored  in  the  magnetic  field  is 

JT       j  •  S*T 

Li^di=    (  Lidi  =  y2LP 

where  I  is  the  value  of  the  current  at  time  T. 
Starting  and  Stopping  Current  in  an  Inductive  Circuit.  —  Re- 

ferring to  equation  (16)  it  is  evident  that  for  t  =  0,  i  =  0  when 
starting  the  current  and  i  =  I  for  t  =  0  in  stopping  the  current, 
since  energy  cannot  be  altered  in  an  infinitely  short  time  and 
therefore  current  cannot  be  established  or  changed  in  an  infinitely 
short  time. 

Thus  when  considering  the  starting  of  a  current  we  have  for 
t  =  0,  i  =  0. 

Substituting  these  values  in  (16) 

0  =  -r  [e  -  C2e°]  =  -i  [e  -  Cj, 


36 
whence, 

and 


ELECTRICAL  ENGINEERING 


(17) 


This  equation  gives  the  value  of  the  current  at  any  instant  after 
the  closing  of  the  switch,  S. 

If  the  impressed  e.m.f.,  e,  is  sud- 
denly short-circuited  by  the  closing  of 
the  switch,  S'  (Fig.  34),  then  e  =  0, 
and,  at  the  instant  of  closing,  t  =  0, 

FIG.  34.  *  =  I, 

where  Z  is  the  current  in  the  circuit 
just  before  closing  the  switch. 
Substituting  these  values  into  (16) 


whence, 
and 


7  =      [0  -  CJ, 


C,  =  -  rZ, 


(18) 


This  equation  gives  the  current  at  any  instant,  t,  as  it  is  dying 
away  in  the  circuit  after  the  e.m.f.  has  been  suddenly  removed. 

The  inductance  of  coils 
varies  with  the  size,  shape 
and  number  of  turns.  If  a 
given  length  of  wire  of  defi- 
nite  size  is  to  be  made  into  a 
coil,  maximum  inductance  FIG<  35. 

will  very  nearly  be  obtained 
if  the  coil  has  the  proportions  given  in  Fig.  35.  1 
the  inductance,  in  this  case,  will  be: 

0.27  cm.2 


The  value  of 


10«Xc 


, 
henrys> 


where  the  length  of  the  coil  is  given  in  centimeters,  and  c  is  in 
centimeters. 

1  BROOKS  and  TURNER,  "Inductance  of  Coils,"  Bulletin  No.  53,  Univ.  of 
Illinois  Engineering  Experiment  Station. 


INDUCTANCE  37 

Problem  26. — Find  and  plot  current  vs.  time  when  the  circuit  is  closed  on 
a  coil  of  1  km.  of  No.  15  B.  and  S.  wire  (diam.,  d.c.c.,  =  0.066  in.;  r/1000'  = 
3.17W),  designed  for  maximum  inductance,  e  =  100  volts. 

Problem  26. — Find  and  plot  the  curve  of  dying  away  of  the  current  when 
the  coil  of  problem  25  is  short-circuited. 

Problem  27. — Find  the  average  value  of  the  inductance  of  the  lifting 
magnet  previously  designed  (Chap.  V),  and  determine  how  long  it  will 
take  for  the  current  to  rise  to  90  per  cent,  of  its  permanent  value. 


CHAPTER  VIII 
ALTERNATING  CURRENTS 

It  has  been  shown  that  the  fundamental  equation  in  an  induc- 
tive circuit  where  the  resistance  and  inductance  are  constants  and 
not  depending  upon  the  current  is  : 


This  equation  gives  the  relation  between  the  particular  values 
of  e.m.f.  and  current  at  any  instant. 

In  the  case  previously  discussed  it  was  assumed  that  the  im- 
pressed e.m.f.,  6,  was  constant. 

In  most  engineering  problems  the  e.m.f.  is,  however,  not  con- 

stant but  it  varies  from  instant  to 
instant.  Almost  all  electrical  instal- 
lations  now  use  alternating  current 
rather  than  direct  current.  In  this 
case  it  will  be  seen  that  the  e.m.f. 

and  current  can  almost  always  be  as- 
FIG.  36.  j-       L         •      i  • 

sumed  to  vary  according  to  a  simple 

sine  law. 

In  other  words  it  can  be  assumed  that  the  instantaneous  value 
of  the  current  at  any  time,  t,  can  be  found  from  equation  i  =  Im 
sin  ut  (Fig.  36),  where  o>  =  27r/  =  angular  velocity,  and  /  = 
frequency  of  alternation  of  the  current  =  number  of  cycles,  or 
complete  reversals,  per  second.  Im  =  maximum  value  of  cur- 
rent. For  60  cycles,  o>  =  27r60  =  377. 

.'.  i  =  Im  sin  377*. 

Differentiating  eqv:  i  =  Im  sin  w 
we  get 

di 

-£  =  7mw  cos  wt. 

Substituting  in  (15), 

e  =  r  Im  sin  wt  +  L  7mw  cos  orf 
=  Im  (r  sin  wt  +  Lw  cos  o>0  (19) 

38 


ALTERNATING  CURRENTS 


39 


Thus  e  is  the  sum  of  two  component  waves,  one  depending  on 
the  sine  of  ut,  and  the  other  on  the  cosine. 


Problem  28.— Let  Im 

ponent  waves  of  e.m.f. 


1,  r  =  0.5,  Leo  =0.4.     Find  and  plot  the  com- 


Sin  at 

Imr  sin  ut  =  ir 
Cos  wt . . 


w  cos 


L* 

dt 


e. 


00 
00 
00 

40 
40 


30° 

.5 

.25 

.866 

.346 
.596 


60° 
0.866 
0.4330 
0.5 


0.2 
0.6330 


90° 
1.0 
.5 
0.0 


0.0 

.5 


120° 
0.866 
0.433 
-0.5 

-0.2 
0.233 


150° 

0.5 

0.25 

-0.866 

-0.346 
-0.096 


180° 
0.0 
0.0 

-1.0 

-0.4 
-0.4 


These  waves  are  shown  plotted  and  combined  in  Fig.  37. 
Problem  29. — A  similar  set  of  waves  should  be  obtained  by  each  student 
from  values  of  Imi  r,  Leo,  assigned  at  random. 

By  inspecting  these  waves  it  is  seen  that  i  lags  behind  e,  that 


FIG.  37. 

is,  it  passes  through  zero  later  than  e  by  about  40°.  This  il- 
lustrates one  of  the  characteristic  features  of  inductive  circuits. 
It  should  also  be  noted  that  ir  is  in  time  phase  with  i,  and  that 
iLu  is  in  time  quadrature  with  i,  being  90°  in  phase  ahead  of  i. 

The  quantity  Lo>  is  called  reactance.  It  is  measured  in  ohms, 
and  denoted  by  the  letter  X.  Thus  Lw  =  X,  where  L  is  the  in- 
ductance in  henrys,  w  =  2ir/  is  the  angular  velocity  in  radians  per 
second,  and  X  is  the  reactance  in  ohms.  X  is  not,  like  L,  a 
property  of  a  coil  or  circuit,  but  depends  on  the  frequency. 

The  average  value  of  the  e.m.f.  generated  in  a  coil  of  a  dynamo, 
depends  only  on  the  speed  of  rotation  and  the  number  of  lines  of 


40  ELECTRICAL  ENGINEERING 

flux  cut;  that  is,  it  depends  on  the  average  rate  of  cutting  of  the 
lines  of  flux,  by  the  conductors,  and  not  on  the  distribution  of 
the  lines  under  the  poles.  The  effective  value  of  e.m.f .  does,  how- 
ever, depend  on  the  distribution  of  the  flux. 

Frequency  has  been  defined  as  number  of  cycles  per  second. 
A  two-pole  generator,  at  1  r.p.s.  has  the  frequency,/  =  1. 

A  four-pole  generator,  at  1  r.p.s.,  has/  =  2. 

fry 

A  p-pole  generator,  at  N,  r.p.s.,  has/  =  -^N. 

The  coil,  in  position  1  (Fig.  38a),  contains  the  whole  flux. 
The  coil,  in  position  2,  contains  no  flux.  Thus,  a  change  of 

the  whole  flux  takes  place  in  a  quar- 
ter  of  a  revolution. 

If  T  is  the  time  of  1  cycle,  the 
whole   flux   is   therefore   cut   in   the 

T 

time  —r  - 
4 

The  average  rate  of  cutting  is  then 
<|>        4$ 
FIG.  38.  ~m"  —  ~7p    where  &  is  the  total  flux. 

T 
Therefore,  the  average  e.m.f.  is  -y-,  where  N  is  the  number  of 

turns,  and  2N  is  the  number  of  conductors  per  circuit. 
At  60  cycles, 


In  general, 


a-.-L 

*        60 


.'.  Average  e.m.f.  =  4N3>f  =  4N$f  X  10~8  volts. 

In    a    four-pole    machine    (Fig.   386)   all  flux  is  cut  in   % 

revolution.     The  average  rate  of  cutting  is  therefore  ~,  where 

I  i 
T\  is  the  time  of  a  revolution. 

Average  rate  =  -^  =  83>Ng  =  ~^-  =  4*/  where  Ns  is  the 

number  of  revolutions  per  second,  and  f  =  -  Ns. 
With  N  turns,  average  rate  =  4/Ar<l>  =  average  e.m.f. 

X  10~8  volts  (20) 


ALTERNATING  CURRENTS  41 

This  equation  is  identical  with  that  for  a  2  pole  machine.  It 
applies  regardless  of  the  number  of  poles  as  long  as  N  is  the 
number  of  turns  in  series  per  circuit. 

Average  Value  of  a  Sine  Wave.  —  The  e.m.f.  induced  by  rota- 
tion of  the  armature  conductors  in  the  field  is 


Let  <p  =  <£m  cos  ut,  be  the  flux  enclosed  at  any  instant.     Then,. 
-T  =  —  o>$TO  sin  ut  is  the  rate  of  change  of  the  flux,  and  e  = 


volts. 


sin  ut. 
In  practical  units, 

sin 


108 

Since  co  =  2irf  this  may  be  written, 

sin  co/ 


10' 

For  maximum  e.m.f., 

sin  u*  =  1,  and  Em  =  ^s'm  volts  (21) 

To  obtain  the  average  value  of  e.m.f.,  integrate  a  half-wave 
and  divide  by  TT,  that  is,  by  the  length  of  a  half-wave. 
Thus, 

««.  =  -  (  "sin  6dd  =  -\  -  cos  0|    =  -  =  0.636. 

irjo  TT|_  Jo  7T 

2  2 

.'.  The  average  value  =  -  X  Em.     Multiplying  (21)  by  -» 

7T  7T 

Av.  e  =  - 


which  agrees  with  the  average  value  previously  found  (20). 

Effective  Value  of  a  Sine  Wave. — Let  i  =  Im  sin  6. 

If  this  current  flows  through  a  resistance  r,  it  has  been  seen 
that  the  heat  developed  at  any  instant  is  i2r. 

Thus,  the  heat  developed  per  cycle  may  be  expressed  as, 

T2% 

r  I    I2m  sin2  Ode. 
Jo 
By  trigonometry, 

sin2  0  =  y>,  —  V<>  cos  20. 


42  ELECTRICAL  ENGINEERING 

Substituting, 

The  average  value  of  energy  flow  or  the  rate  at  which  energy 
is  being  dissipated,  or  the  power,  is 

rlmir  =  rim 

27T  2 

r/i 
Thus,  the  rate  of  heat  dissipation  is  -£-• 

The  effective  value  of  the  current  corresponds  to  a  constant 
or  direct  current  which  would  give  the  same  heat  in  the  same 
time  if  flowing  through  the  same  resistance. 


.  =       ,  and  iett.  =          =  0.707  Im. 


Similarly,  the  effective  value  of  e.m.f.  is  obtained,  and  eeff.  = 
0.707  Em,  where  Em  is  the  maximum  value  of  the  sine  wave  of 

electromotive  force. 

effective  value  .        „    ,  .  ,      - 

I  The  ratio  -  —  is  called  the  form 

N  average  value 

factor. 

j  With  sine  waves  form  factor  (j(f)  =  QQ^Q  = 

^_  —  .  — 


. 

The  equation  for  the  effective  value  of  the 

I  e.m.f.  is  obtained  from  (21)  by  multiplication 

i 

Fro.  39.  ' 


m          .m 
-       '"~    V210«  ^0^^ 

This  applies  to  a  concentrated  coil  of  N  turns. 

If,  however,  the  turns  are  distributed  over  the  periphery,  as 
in  a  direct-current  armature,  from  Fig.  39  it  is  seen  that  coil  1 
contains  all  the  flux,  while  coil  2  contains  the  flux  X  cos  6. 
Therefore  the  .effectiveness  of  coil  2  is  Nc  cos  0,  where  Nc 
number  of  turns  of  the  coil. 

Let  N  =  total  number  of  turns.     Then  the  turns  per  cm.  of 

N 
armature  periphery  =  ^  —  ,  where  r  =    radius  of  armature. 


ALTERNATING  CURRENTS  43 

N 
Then  the  effectiveness  of  the  turns  per  cm.  is  ~ —  cos  6.     The 

average  effectiveness  of  the  turns  per  cm.  is 


NcosB..        N 


r+i 

1    I  N  cot 
,J      l.r 


N       2 
The  total  effectiveness  is  therefore  2rr  X  -^  =  -N. 

Therefore,   in  a  distributed  winding,  the   turns   are   not  so 
effective  as  when  they  are  concentrated. 
Thus,  for  distributed  winding, 

4  44fAT<iv       2 

*±»^*//>  **m          ^ 


CHAPTER  IX 


DIRECT -CURRENT  GENERATORS 

Homopokr  Generators. — These  are  also  called  by  the  names 
" acyclic"  and  "unipolar."  They  are  a  small  class  of  machines, 
distinguished  from  the  usual  types  of  direct-current  machinery 
in  that  the  conductors  always  move  through  the  magnetic  field 
in  a  constant  direction  with  respect  to  the 
direction  of  the  lines  of  flux. 

Among  the  earliest  of  dynamos  may  be 
mentioned  one  of  this  type  known  as 
" FARADAY'S  Disc  Dynamo,"  in  which  a 
copper  disc  was  rotated  between  the  poles 
of  a  permanent  magnet. 

Current  was  collected  by  means  of  two 
brushes  making  contact,  respectively,  with  the  rim  and  axle  of 
the  disc  (Fig.  40) .  A  more  modern  type  of  homopolar  generator 
is  shown  diagrammatically  in  Fig.  41.  For  the  permanent  mag- 


FIG.  40. 


I, 

\ 

;                    i 

! 

f 

\ 

< 

i 

N 

E 

3 

r 

p 

1         F 

^ 

- 

1 

Q 

f 

L_ 

^ 

,  > 

^ 

) 

^ 

V 

( 

f. 

~""\ 

J 

\ 

_s 

FIG.  41. 

net  is  substituted  a  powerful  electromagnet,  and  two  sets  of 
brushes  are  used  instead  of  one.  By  connecting  these  brushes 
in  series  outside  of  the  machine  the  total  e.m.f.  at  the  terminals 
is  doubled. 

44 


DIRECT-CURRENT  GENERATORS 


45 


From  the  fundamental  considerations  developed  in  Chap.  VI 
it  is  evident  that  the  voltage  between  each  set  of  brushes  is 


e  — 


108 


where  N  =  revolutions  per  second  and  3>  =  total  flux,  since 
there  is  only  one  conductor  between  the  brushes.  With  any  ar- 
rangement which  permits  the  use  of  additional  sets  of  brushes, 
as  in  Fig.  41,  the  voltage  is  increased  in  proportion  to  the  number 
of  sets  of  brushes  connected  in  series,  and  becomes 


e  = 


108 


FIG.  42. 

where  c  is  the  number  of  conductors  and  is  equal  to  the  number 
of  sets  of  brushes  in  series.  Fig.  42  illustrates  the  use  of  bar 
conductors  on  the  armature.  Each  conductor  is  connected  to 
two  slip  rings  on  which  brushes  bear.  There  are  thus  twice  as 
many  slip  rings  as  conductors.  Since  the  conductors  are  in- 
sulated, they  may  be  put  in  series  by  properly  connecting  their 
brushes  outside  of  the  machine. 

Direct-current  Machines  with  Commutators. — On  most 
direct-current  machines  use  is  made  of  commutators.  To 
understand  these  machines  a  knowledge  of  the  principle  of  wind- 
ings on  the  armature  is  needed.  In  Fig.  43  a  single  coil  is  repre- 
sented in  a  magnetic  field.  The  ends  of  the  coil  are  connected 
to  the  segments  of  a  two-part  commutator.  In  the  position 


46 


ELECTRICAL  ENGINEERING 


shown,  the  e.m.f.  is  maximum.  As  the  coil  moves  in  the  field, 
the  segments  move  under  the  brushes  and  the  e.m.f.  at  the 
brushes,  AB,  during  a  half  revolution,  has  the  values  of  a  half- 
sine  wave.  When  this  e.m.f.  reaches  zero,  the  segments  pass 
from  under  the  brushes.  The  same  operation  is  then  repeated 

and  gives  a  succession  of  half  waves, 
all  in  the  same  direction.  If  now 
another  loop  is  placed  on  the  armature 
at  90°  to  the  first  one,  a  new  series  of 
half  waves  will  be  added  at  90°  to  the 
first  series.  By  connecting  these  loops 
in  series,  suitably  joining  to  commuta- 
tor segments  and  continuing  to  use 
only  two  brushes,  the  e.m.fs.  of  the 
loops  are  added  together  and  produce 
a  resultant  e.m.f.  shown  in  heavy  dots  by  the  wave  "d"  (Fig. 
44).  This  wave  never  reaches  zero  and  is  much  more  steady 
than  that  produced  by  a  single  coil.  By  continuing  this  process, 
all  irregularities  are  virtually  wiped  out  and  there  results  a 
smooth  wave  of  constant 
e.m.f.  A  simple  example 
of  armature  winding  with 
commutator  and  brushes 


FIG.  44. 


is  shown  in  Fig.  45,  for  the  purpose  of  illustrating  the  connec- 
tion of  coils  in  series. 

Types  of  Direct-current  Commutator  Machines. — Direct- 
current  machines  are  usually  divided  into  groups  according  to 
the  method  of  exciting  the  field  magnets,  as  follows: 

1.  Permanent  Magnet  Machines. — These  have  no  field  windings, 
but  the  field  structure  consists  of  hard-steel  permanent  magnets. 
They  constitute  a  small  group,  used  chiefly  for  telephone  sig- 
nalling and  gas  engine  ignition. 


DIRECT-CURRENT  GENERATORS 


47 


2.  Separately  Excited  Machines. — In  these,  the  field  winding 
is  supplied  with  current  from  an  external  source.     The  chief 
advantage  of  this  type  is  that  it  enables  a  steady  field  excitation 
to  be  maintained  at  all  times  regardless  of  the  fluctuations  in 
voltage  at  the  brushes. 

3.  Shunt  Machines. — In  this  type  the  source  of  excitation  of 
the  field  is  derived  from  the  terminals  of  the  machine  itself. 
The  field  circuit  is  connected  in  parallel  with  the  external  cir- 
cuit and  the  field  current  varies  as  the  voltage  of  the  machine 
changes. 

4.  Series  Machines. — The  current  in  the  armature  is  made  to 
flow  also  through  the  field  windings;  that  is,  the  field  and  armature 
coils  are  connected  in  series  with  the  external  circuit.     Thus  the 
field  excitation  is  proportional  to  the  load  current. 


1  Permanent  Magnet    2  Separate  Excitation        3  Series 

FIG.  46. 


4  Siunt 


5  Compound 


5.  Compound  Machines. — These  are  excited  partly  by  a  shunt 
winding  and  partly  by  a  series  winding,  each  pole  being  provided 
with  both  a  shunt  and  series  coil.  The  total  field  excitation 
thus  depends  upon  the  voltage  of  the  machine  as  well  as  on  the 
load  current. 

These  five  types  are  illustrated  in  Fig.  46.  Other  com- 
binations are  sometimes  used  in  special  cases.  The  performance 
characteristics  of  these  various  types  of  generators  differ  greatly. 
In  general,  the  characteristic  of  a  generator  is  a  curve  showing 
the  relation  between  terminal  voltage  and  the  load  current,  the 
latter  being  the  independent  variable.  These  curves  and 
others  of  a  similar  nature  should  be  thoroughly  studied,  especially 
in  the  laboratory. 

Armature  Reaction. — When  a  generator  is  delivering  no  cur- 
rent the  direction  of  the  field  flux  is  along  the  axis  of  the  poles. 


48 


ELECTRICAL  ENGINEERING 


When  current  is  flowing,  however,  the  armature  becomes  an 
electromagnet  on  its  own  account,  and  the  field  flux  becomes  the 
resultant  of  that  produced  by  the  field  windings  and  that  due 
to  the  armature  winding. 

Fig.  47a  shows  a  bipolar  dynamo  with  a  ring  armature.  Arrows 
show  the  direction  of  current  and  also  of  flux.  Starting  from 
the  negative  brush,  the  current  divides  as  it  enters  the  armature, 
one  half  winding  around  to  the  left,  the  other  half  pursuing  a 
similar  path  to  the  right,  and  both  finally  joining  again  to  enter 
the  positive  brush.  It  is  to  be  noted  that  the  flux  set  up  by 
these  armature  currents  is,  in  general,  in  space  quadrature  to  the 
flux  due  to  the  field  winding. 


FIG.  47. 

Fig.  476  shows  an  equivalent  diagram  representing  a  drum 
armature.  When  once  the  principles  of  current  action  in  the 
armature  are  understood,  it  is  simpler  to  make  use  of  the  rep- 
resentation of  Fig.  476  than  of  Fig.  47a.  For  clearness,  the  com- 
mutator is  omitted  in  the  case  of  the  drum,  the  position  of  the 
brushes  being  indicated  with  reference  to  the  armature  itself. 
It  makes  no  difference  how  the  end  connections  are  made,  so 
far  as  the  armature  m.m.f.  is  concerned.  In  this  case,  since  the 
brushes  are  not  shifted  but  are  placed  on  the  so-called  neutral 
axis  midway  between  the  poles,  the  armature  magnetomotive 
force  is  directed  vertically  upward,  while  the  field  magnetomo- 
tive force  is,  as  always,  along  the  pole  axis.  The  resultant 
magnetomotive  force  is  the  vector  sum  of  these  two.  Since  the 
armature  m.m.f.  acts  at  right  angles  to  the  field  m.m.f.,  its  effect 
is  said  to  be  wholly  cross-magnetizing. 

When  the  brushes  are  shifted  a°  the  armature  m.m.f.,  still 
acting  along  the  brush  axis,  may  be  resolved  into  components. 


DIRECT-CURRENT  GENERATORS 


49 


pc  =  pA  cos  a,  the  cross-magnetizing  component,  acting  at 
right  angles  to  the  field,  and  FD  =  FA  sin  a,  the  demagnetizing 
component,  acting  directly  in  opposition  to  the  field.  The  re- 
sultant m.m.f.,  OR,  Fig.  48,  is  then  due  to  the  m.m.fs.  of  the 
field  OF  and  the  armature  OA,  the  latter  being  composed  of  OC, 
cross-magnetizing,  and  OD,  demagnetizing. 

Cross-magnetization  is  always  present  when  the  armature 
carries  current.  It  distorts  the  field  and  displaces  the  neutral 
axis,  necessitating  thereby  a  shifting  of  the  brushes.  When  the 
brushes  are  shifted,  demagnetization  also  enters  in,  weakening 
directly  the  field  strength.  Under  such  conditions  the  resultant 
flux  takes  up  a  general  direction  as  indicated  by  the  shading  in 
the  air  gap  in  the  figure.  The  pole  tips  are  unequally  magnet- 
ized, the  leading  tips  being  weakened  and  the  trailing  tips 
strengthened. 


A   c 


FIG.  48. 


The  actual  direction  of  the  resultant  flux  is  not  along  OR  but 
along  OR' ',  a  line  of  somewhat  less  deviation  from  OF.  This  is 
because  of  the  unequal  reluctances  of  the  paths  along  the  direc- 
tions of  the  component  m.m.fs. 

Consider,  for  example,  a  generator  whose  flux  per  pole  enter- 
ing the  armature  is  <£r,  under  conditions  of  normal  operation, 
that  is,  voltage  and  speed. 

To  generate  this  flux  at  no  load  would  require  F0  amp. -turns 
on  the  field  core  if  all  the  flux  generated  in  the  field  passed  through 
the  armature.  Some  flux,  however,  passes  around  the  armature 
without  cutting  its  conductors.  This  is  called  leakage  flux, 
and  amounts  to  15  or  20  per  cent,  of  the  net  flux,  in  ordinary 
machines.  To  provide  this  leakage  flux  as  well  as  the  net  flux, 
</v,  requires  kF0  field  amp.-turns,  where  fc  is  the  leakage  co- 
efficient and  may  be  taken  as  1.15. 

Let  Ic  amp.  now  flow  in  each  armature  conductor,  and  let  the 


50  ELECTRICAL  ENGINEERING 


total  number  of  conductors  be  C.     Then  total  turns  =  -a  and 

turns  per  pole  =  ~~>  where  p  =  number  of  poles. 

CIe 

The  total  armature  amp.-turns  per  pole  =  -~~' 

Let  the  brushes  be  on  the  geometrical  neutral,  that  is,  midway 
between  the  poles.  Then,  since  the  conductors  are  distributed 

over  the  entire  periphery,  the  effective  armature  A.T.  per  pole 

9      r       T  r 
=  £  /  ±L  =  ffk. 

TT    c  2p         irp 

The  effect  of  these  distorting  ampere-turns  has  been  shown 
to  be  to  weaken  the  flux  in  the  leading  pole  tips  and  to  strengthen 
that  in  the  trailing  tips.  The  net  result  owing  to  unequal 
saturation  of  the  iron,  is  to  reduce  the  actual  amount  of  the  flux. 
In  order  to  compensate  for  this  reduction  extra  ampere-turns 
must  be  placed  upon  the  field  core  to  the  amount  of  about  40 
per  cent,  of  the  armature  cross-magnetizing  ampere-turns. 
Thus, 

Fc  =  — — — >  if  there  is  no  brushshift 
irp 

and  the  total  field  amp.-turns  per  pole  are 
Ft  =  kF.  +  °-^. 

vp 

When  the  brushes  are  shifted  a°,  the  cross-magnetizing  amp.- 
turns  are  Fc  =  — ~  X  -j~  (Fig.  49). 

.    180  -  2«  ICC 
Their  effective  value  is  kc  — r^ —  TT~ 

loU  4p 

where 


r* 

Ur 


2  cos  a 
cos  0  dO,   = 


7T    — 


where  0°  =  IT  —  2a 

The  demagnetizing  turns   consist  of  a  belt  of  conductors  of 
width  2a°.     The    effective  demagnetizing  amp.-turns  are  then 

,      2a    ICC 


DIRECT-C  URRENT  GENERA  TORS  51 

where 

/»+! 

2  sin  a 

cos  ede  =  ~2^- 


These  latter  ampere-turns  act  in  direct  opposition  to  the  field. 
If  there  were  no  leakage  of  flux  between  field  and  armature,  they 
would  be  compensated  by  placing  an  equal  number  of  additional 
ampere-turns  on  the  field.  Owing  to  leakage  this  number  must 
be  multiplied  by  k. 

The  total  required  field  ampere-turns  under  the  condition  of 
brush  shift  of  a°  and  Ic  amp.  in  the  armature  conductors  is 
then 

(90  -  a°)IcC 


/ 

( 


0.4 


in  order  that  the  flux  entering  the  armature  shall  be  <pr.  With 
constant  generated  voltage  the  terminal  voltage  falls  off  as 
the  load  increases,  due  to  the  IR  drop  of  the  armature.  <pr 
must  therefore  be  increased  sufficiently  to  make  up  for  the  IR 
drop. 

CHARACTERISTICS  OF  DIRECT  CURRENT  GENERATORS 

From  the  discussion  given  above  it  should  be  possible  to  calcu- 
late the  change  of  voltage  with  load  in  any  of  the  different  types, 
provided  that  the  saturation  curve  could  be  expressed  in  a  simple 
manner.  This  is  unfortunately  not  possible,  but  it  can  be  approx- 
imated by  FROELICH'S  equation,  which  is  : 

km     2 


where  m  is  the  excitation  in  ampere-turns  and  e  the  corresponding 
voltage,  k  and  ki  are  constants  depending  upon  the  shape  of  the 
saturation  curve  which  constants  can  be  determined  by  substi- 

F  * 

1  It  is  seen  that  jf  =  cot  a.    Thus,  on  the  basis  given  above,  the  ratio 

between  the  actual  ampere-turns  needed  to  compensate  for  the  cross-mag- 
netizing and  demagnetizing  ampere-turns  is  .34  cot  a. 

2e  is  the  induced  e.m.f.  due  to  the  rotation  of  the  armature  conductors  in 
the  magnetic  field,  m  is  the  resultant  m.m.f.  of  the  amp.-turns  on  field 
and  armature  referred  to  the  field  structure.  At  no-load,  m  is  obviously 
the  field  excitation  alone. 


52  ELECTRICAL  ENGINEERING 

tuting  two  known  values  of  e  and  m  from  the  actual  saturation 
curve. 

Consider  a  compound  wound  generator.     Let  the  terminal 
voltage  be  e  and  the  load  current  i.     If  the  resistance  of  the  shunt 

p 

field  winding  is  rf  the  shunt  field  current  is  if  =  —.l  If  each  field 
spool  has  t  turns  then  the  m.m.f.  of  the  field  winding,  per  pole, 

/j 

is  m\  —  i/t  =  —t.     If  the  series  winding  has  ti  turns,  per  pole, 

the  m.m.f.  per  pole  of  the  winding  is  mz  =  ii\.  Let  the  demag- 
netizing ampere-turns  per  pole  of  the  armature  with  full-load 
current  as  determined  above  be  D;  then  the  demagnetizing  ampere- 

turns,  with  load  current  it  is  ms  =  ~ji,  where  7  is  full-load  current. 

Let  the  equivalent  demagnetizing  ampere-turns  with  full-load 
current,  due  to  the  "  cross  magnetizing"  ampere-turns  be  C; 

C 
then  the  demagnetizing  effect  of  i  amp.  is  w4  =  ~^i. 

The  total  m.m.f.,  m0,  on  each  pole  —  were  there  no  leakage  — 
is: 


Due  to  the  leakage  between  the  field  poles  the  equation  is 
obviously  modified.  Assuming  15  per  cent,  leakage: 

w0  =  0.85(wi  +  7/12)  —  w3  —  w4, 

ra0  then  being  the  m.m.f.  which  causes  flux  to  interlink  with  the 
armature  conductors. 

If  the  saturation  curve  were  plotted  on  the  basis  of  these 
ampere-turns  the  corresponding  voltage  could  be  obtained  either 
directly  or  through  FROELICH'S  equation.  This  is,  however,  not 
the  case  but  the  saturation  curve  takes  into  consideration  the 
leakage,  therefore,  in  order  to  use  the  saturation  curve  we  have 
to  use  a  new  value  of  m0,  namely: 


ra3 
,  +  «,-   - 

Numerical  Application.  —  Let  the  no-load  voltage  e0,  the  no- 
load  excitation,  and  the  full-load  current  be  taken  as  unity,  then 

mi  =  e 

C~*lT 

1  It  is  really  —  ,  but  ir  is  usually  very  small. 


DIRECT-CURRENT  GENERATORS 


53 


Let  the  full-load  series  excitation  be  40  per  cent,  of  the  no-load 
excitation,   then 

ra2  =  OAi 

Let  the  demagnetizing  ampere-turns  of  the  armature  with  full- 
load  current  be  10  per  cent,  of  the  no-load  field  excitation,  then 
ra3  =  0.1  Oz,  and  let  the  equivalent  demagnetizing  ampere-turns 
of  the  cross-ampere-turns  with  full-load  current  be  20  per  cent., 
then  m4  =  0.20i. 

(This  relation  between  ra3  and  ra4  corresponds  to  11°  brush 
shift.) 

1.2 


1.0 


,0.8 


'0.6 


0.4 


0.2 


0.2 


0.4 


0,6         0.8          LO 
Current 

FIG.  49. 


1.2 


1.4         1.6 


Then 


Referring  now  to  FROELICH'S  equation  and  assuming  the  sat- 
uration curve  to  be  such  that  for  e  =  1,  m  =  1;  for  e  =  0.6,  m  = 
0.5;  then  it  is  readily  proven  that  k  =  1.5,  and  ki  =  0.5 


1.5m 
. .  e  =  ..    .  ~  , —  or  m  = 


'  1.5  -  0.5e 


0.5  mv         ~  1.5-  0.5  e 

=  e  +  0.05i,  or  e  =  0.5  -  0.025z  + 

V(0.5  -  0.025t)2  +  0.: 


The  voltage  at  the  terminal  of  the  machine  is  less  than  e  by 
the  ir  drop  in  the  armature  winding,  brushes  and  series  field. 

If  at  full-load  the  drop  is  3  per  cent,  then  at  any  other  load  it  is 
0.03i. 


54  ELECTRICAL  ENGINEERING 

Thus  e»  the  terminal  voltage  is 


€l  =  e  -  0.03*'  =  0.5  -  0.055i  -f  V(0.5  -  0.025^) 2  +  0.15* 

The  student  should  verify  curves  a  and  6,  Fig .,.49.  Curve  a 
applies  to  the  compound  wound  generator  discussed  above. 
Curve  b  to  a  typical  shunt  generator  in  which  the  ratio  between 
the  armature  reaction  and  the  no-load  excitation  is  less  than  with 
a  compound  wound  generator,  and  in  which  the  saturation  at 
normal  voltage  is  usually  higher. 

The  constants  used  for  the  shunt  generator  are: 

k  =  2.33,  fcj  =  1.33,  mi  =  e,  ra2  =  0,  w3  =  O.OSz,  m4  =  O.lOi, 
Ir  =  3  per  cent. 

It  is  seen  that  as  the  resistance  of  the  load  is  gradually  decreased 
the  current  increases  up  to  a  certain  maximum  value,  in  this 
case  20  per  cent,  more  than  rated  current;  after  that,  the  current 
and  voltage  both  decrease. 

The  student  should  study  the  effect  of  the  saturation  on  the 
shape  of  these  curves. 


CHAPTER  X 

A    STUDY    OF    THE    DESIGN    OF    A    DIRECT-CURRENT 

GENERATOR 

All  the  underlying  principles  of  the  direct-current  generator 
may  be  studied  to  good  advantage  from  the  basis  of  a  concrete 
example.  The  example  here  chosen  is  an  ordinary  compound- 
wound  generator  with  the  following  specifications: 

M.P.  12  -  500  -  375  -  250  volts,  which  means  that  it  be- 
longs to  the  general  multipolar  class  (M.P.),  has  12  poles,  500 
kw.  rated  output,  375  r.p.m.  at  normal  speed,  and  the  voltage 
is  250  at  both  no-load  and  full-load.  The  normal-load  current 
may  also  be  given.  It  is 

=  2000  amp. 

Other  data  for  this  machine  are:  Armature  external  diameter  = 
64  in.,  from  which  is  obtained  what  is  called  diameter  per  pole 

64 
=  12  =  5-33  in- 

Armature  internal  diameter  =  44  in. 

Number  of  armature  slots  =  216. 

Dimensions  of  slots,  0.465  in.  wide  by  1.3  in.  deep. 

Armature  winding  is  of  the  multiple  drum  type. 

Current  in  each  effective  conductor  is 

^    -     ,  /.-??£-  167  amp.'      , 

Each  effective  conductor  consists  of  two  bars  in  parallel.  Each 
bar  is  0.075  in.  X  0.45  in.,  without  insulation. 

Area  of  each  effective  conductor  =  2  X  0.075  X  0.45  = 
0.0675  sq.  in. 

.'.Current  density  in  conductor  =  Q  0675  =  24^°  amp*  per 
sq.  in. 

55 


56  ELECTRICAL  ENGINEERING 

With  direct-current  generators,  current  density  in  the  arma- 
ture conductors  generally  lies  between  2000 
and  3000  amp.  per  sq.  in. 

Number  of  effective  conductors  per  slot  =  4. 

Number  of  conductor  bars  in  each  slot  =  8. 

Arrangement    of    conductors    in    slot   is   as 


y  ,  .         T,. 

FIG.  50.  shown  in  Fig.  50. 

Number  of  effective  turns  per  pole, 

_  conductors  per  slot  X  number  of  slots  _  4  X  216  _ 
conductors  per  turn  X  poles  2  X  12 

Flux  Calculation. — There  are  now  sufficient  data  to  apply  the 
fundamental  e.m.f.  equation  to  the  determination  of  the  flux. 
The  equation  is 

*  =  %<  volts.  • 

Supplying  numerical  values, 

4  X  37.5  X  0a  X  36 

"lO5"" 
whence 

0a  =  4,630,000. 

<£a  here  is  the  required  flux  per  pole  entering  the  armature  at  no- 
load.  Neglecting  the  effect  of  the  small  shunt  field  current  flow- 
ing in  the  armature,  the  generated  voltage  and  terminal  voltage 
are  the  same  at  no-load. 

At  full-load,  in  order  to  maintain  the  same  terminal  voltage, 
250,  it  would  be  necessary  to  generate  a  slightly  higher  voltage 
to  supply  the  drop  in  the  armature,  series  field  and  brushes. 
Assuming  this  drop  to  be  2^£  per  cent.,  the  required  flux  entering 
the  armature  at  full-load  is 

0'a  =  1.025  X  4,630,000  =  4,750,000. 

The  total  flux  which  must  be  generated  is  made  up  of  the 
armature  flux  and  that  which  leaks  across  from  pole  to  pole  with- 
out passing  through  the  armature.  Assuming  the  leakage  flux 
to  be  15  per  cent.,  the  total  flux  in  the  pole  core  at  no-load  will  be 

0C  **  1.15  X  4,630,000=  5,320,000. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    57 


At  full-load  the  total  flux  will  be 

0'c  =  1.15  X  4,750,000  =  5,450,000. 

The  Magnetic  Circuit. — In  order  to  produce  this  flux  it  is 
necessary  to  employ  the  required  number  of  ampere-turns  per 
pole  of  the  field  winding.  These  are  determined  as  the  sum  of 
the  ampere-turns  required  for  each  part  of  the  magnetic  circuit 
supplied  by  the  windings  on  a  single  pole.  The  separate 
parts  are  (1)  the  armature  teeth,  (2)  the  air  gap,  (3)  the  armature 
core,  (4)  the  pole  core,  (5)  the  yoke. 

The  relations  between  ampere-turns  per  inch  length  of  the 
magnetic  path  and  flux  density  in  lines  per  square  inch  are 
given  by  the  saturation  curves  for  the  various  materials  com- 


.    FIG.  51. 

posing  the  magnetic  circuit.  To  ascertain  the  ampere-turns  it 
is  necessary  to  know  the  cross-sectional  area  and  length  of  each 
component  part  of  the  magnetic  circuit.  These  are  best  deter- 
mined with  the  help  of  a  scale  drawing  showing  the  armature  and 
the  field  cores  in  their  relative  positions.  Such  a  drawing  is 
reproduced  in  Fig.  51. 

Here  the  mean  flux  paths  are  indicated  by  heavy  dotted  lines. 
The  cross-sectional  areas  through  which  they  pass  are  ascertained 
directly  from  the  given  dimensions,  except  in  the  cases  of  teeth 
and  gap. 

Area  of  Flux  Path  through  Teeth. — Since  the  slot  is  of  uniform 
width,  the  tooth  must  be  narrower  at  the  base  or  "root"  than 


58  ELECTRICAL  ENGINEERING 

at  the  face.  The  ampere-turns  required  for  the  teeth  may  be 
taken  as  the  mean  of  the  ampere-turns  which  would  be  required 
if  the  teeth  area  throughout  were  that  at  their  face,  and  if  it 
were  that  at  their  base.  This  is  not  the  same  as  the  ampere- 
turns  required  for  the  mean  area  of  the  teeth. 
Width  of  tooth  at  face  =  slot  pitch  —  slot  width 

TT  X  64 


216 


-  0.465  =  0.465  in. 


Width  of  teeth  at  face  =  0.465  in.  X  number  of  teeth  under  one 

pole  =  0.465  X  teeth  per  pole  X     P?le  *f  C.    X  1.08 

pole  pitch 

216 
=  0.465  X  -jg-  X  0.72  X  1.08  =  6.53  in. 

The  factor  1.08  is  inserted  to  allow  for  fringing,  that  is, 
the  spreading  of  the  flux  to  teeth  not  immediately  under  the 
pole. 

Teeth  area  at  face  =  6.53  X  net  armature  length  =  6.53  X 
(gross  armature  length  —  air  duct  width)  X  0.9.  This  armature 
has  a  gross  length,  parallel  to  the  shaft,  of  9  in.;  the  air  ducts 
are  six  in  number,  each  %  in.  wide,  making  a  total  width  of  air 
duct  of  2.25  in.  The  factor  0.9  is  commonly  used  to  allow 
for  space  lost  between  the  laminations  due  to  the  presence 
of  Japan  insulation  or  natural  inequalities  in  the  material. 
Substituting  these  numerical  values,  the  teeth  area  at  face 
is  =  6.53  X  (9  -  2.25)  0.9  =  6.53  X  6.07  =  39.6  sq.  in. 

Width  of  tooth  at  base  =  *          A  ~  0.465  =  0.428  m. 


216 
Teeth  area  at  base  =  0.428  X  -jg-  X  0.72  X  1.08  X  6.07  = 

36.4  sq.  in. 

Area  of  Flux  Path  through  Gap.  —  It  might  be  assumed  that  a 
mean  area  between  that  of  the  pole  face  and  that  of  the  teeth 
should  be  taken  for  the  gap.  .  Consideration,  however,  will  show 
that  this  will  give  too  small  a  result. 

The  flux  in  the  gap  fills  practically  the  whole  of  it,  though 
near  the  teeth  the  distribution  is  no  longer  uniform.  A  fairly 
satisfactory  approximation  to  the  effective  gap  area  is  obtained 
by  taking  one-fourth  of  the  sum  of  three  times  the  pole  face  area 
plus  the  teeth  area.  Thus,  gap  area, 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    59 
_3  X  pole  face  area  +  teeth  area 


Pole  face  area  =  pole  arc  X  pole  length  =  12.2  X  7.25  =  88.4 
sq.  in. 

.     A        3  X  88.4  +  39.6       _ 
.  .  AO  =  -        —  -j—         -  =  76  sq.  in.  approx. 

Areas  of  Armature  Core,  Pole  Core  and  Yoke.—  The  armature 
core  section  perpendicular  to  the  flux  path  is  taken  radially, 
and  is  the  product  of  the  radial  distance,  a,  in  Fig.  51  and 
the  net  length  of  the  core.  Thus 

Aa  =  8.7  in.  X  6.07  in.  =  52.8  sq.  in. 

The  pole  core  section  is  circular  in  this  machine,  the  area 
being 

Ap  =  irrp*  =  TT  X  (4.4375)  2  =  62  sq.  in. 

The  effect  of  any  variation  due  to  the  pole  shoe  is  very  slight 
and  may  be  neglected. 

The  yoke  section  is  taken  radially  as  at  6,  in  the  figure.  Its 
form  is  somewhat  irregular. 

In  this  instance  the  area  is 

Ay  =  83.5  sq.  in. 

Materials.  —  The  armature  core  is  of  soft  sheet-iron  laminations 
of  high  permeability,  the  poles  are  of  soft  steel  and  the  yoke  is 
of  cast  iron.  Magnetization  curves  of  these  materials  are  given 
in  Fig.  20. 

No-load  and  Full-load  Saturation  Curves.  —  Having  now  de- 
termined the  fluxes,  areas,  lengths  and  materials,  it  is  in  order 
to  put  these  together  in  tables  to  show  the  flux  densities, 
ampere-turns  per  inch,  and  ampere-turns  for  each  part  of  the 
magnetic  circuit,  and  finally  the  total  ampere-turns.  This  is 
done  for  both  no-load  and  full-load.  In  the  former  case  a 
point  is  obtained  on  the  no-load  saturation  curve,  Fig.  52. 
Other  points  on  this  curve  are  obtained  by  repeating  the  tabu- 
lation process,  starting  with  any  desired  values  of  voltage,  such 
as  80,  150,  200,  260,  280,  determining  the  fluxes  from  the  e.m.f. 
equation,  then  flux  densities  and  ampere-turns. 

With  the  rest  of  the  design  the  student  should  hand  in  both 
curves  with  complete  tabulation  of  points. 


60 


ELECTRICAL  ENGINEERING 


Tabulations  for  250  volts,  no-load  and  full-load,  are  given 
in  Tables  III  and  IV. 

In  the  case  of  full-load  there  must  be  added  the  ampere- 
turns  required  to  overcome  the  armature  reaction,  in  order  to 
give  the  total  required  ampere-turns  and  the  resultant  point 
on  the  full-load  saturation  curve. 

TABLE  III 


Part 

No-load.     E  =  250  volts 

Flux  (mgl.) 

Area 

B 

A.T./in. 

Length,  in. 

A.T. 

Teeth  
Gao 

4.63 
4.63 
2.315 
5.32 
2.66 

uired  amp.-tur] 

(face)  39.  6 
(base)  36  .  4 
76.0 
54.5 
62.0 
83.5 

as. 

117,000 
127,000 
61,000 
42,500 
86,000 
32,000 

£}»» 

19,100 
3 
40 
50 

1.3 

0.3125 
7.0 
12.0 
12.0 

429 

5,970 
21 
480 
600 

7,500 

Armature.  . 
Pole   .  .  . 

Yoke 

Total  req 

TABLE  IV 


Part 

Full-load.     E  =  250  volts 

Flux  (mgl.) 

Area 

B 

A.T./in. 

Length,  in. 

A.T. 

(face)  39.6 
(base)  36.  4 
76.0 
54.5 
62.0 
83.5 

120,300 
130,700 
62,600 
43,600 
88,000 
32,800 

230  \ 
550  /  39° 
19,600 

2.8 
43 
52 

1.3 

0.3125 
7  .0 
12.0 
12.0 

507 

6,140 
19.6 
516 
624 

Teeth  
Can 

4.75 
4.75 
2.375 
5.45 
2.725 

ns  

Armature.  . 
Pole  
Yoke  

Amp.-tur 
Amp.-tur 

Total  req 

7,807 
2,620 

ns  required  tc 
uired  amp.-ti 

>  overcome  armature  r 
jrns 

eaction  

10,427 

Armature  Reaction. — " Armature  reaction"  means  effective 
ampere-turns  per  pole  on  the  armature.  The  actual  amp.- 
turns  per  pole,  in  this  case,  are  167  X  36  =  6000. 

Since  the  turns  are  distributed  over  the  armature  surface  the 
effective  amp.-turns  are 


-  X  6000  =  3820. 

7T 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    61 

If  there  were  no  shift  to  the  brushes,  these  ampere-turns  would 
all  be  cross-magnetizing,  or  distorting.  To  compensate  for 
them,  it  is  necessary  to  supply  about  40  per  cent,  of  their  value 
in  additional  ampere-turns  on  the  field  core. 

It  is  assumed,  however,  that  the  brushes  will  be  shifted  15°, 
giving  a  distorting  belt  of  180°  -  30°  =  150°.  To  overcome  the 
distorting  ampere-turns  at  full-load  there  will  then  be  required 


240 


200 


.120 


40 


7 


2000 


4000 


6000          8000 
Ampere-Turns 

FIG.  52. 


10000 


12000 


150 
180 


X  36  X  167  X  ke  X  0.4  =  1480, 


where 


+  2 


kc  =  ~    \   cos  ed0  =  x^  [1.9318]  =  0.737. 


The  demagnetizing  ampere-turns  constitute  a  belt  30°  wide. 
To  compensate  for  them  would  require  their  exact  numerical 
equivalent,  were  there  perfect  mutual  induction  between  these 
turns  and  the  field.  Owing  to  magnetic  leakage  there  should  be 
added  about  15  per  cent,  to  the  effective  demagnetizing  ampere- 


62  ELECTRICAL  ENGINEERING 

turns.     To  compensate  for  these,  therefore,  will  require 

30 

X  36  X  167  X  fed  X  1.15  =  1140  amp.-turns, 


where  kd  =  0.99. 

To  overcome  armature  reaction  at  full-load  will  require  1480 
+  1140  =  2620  additional  amp.-turns  on  the  field  core. 

For  any  other  load,  keeping  the  same  shift,  the  required 
ampere-turns  will  be  proportional  to  the  load  current. 

No-load  and  full-load  saturation  curves  are  shown  in  Fig.  52. 

The  Shunt  Field  Winding.  —  Under  no-load  conditions  it  is 
evident  that  the  shunt  field  current  must  supply  the  entire  ex- 
citation. In  this  machine,  therefore,  the  shunt  field  m.m.f. 
must  consist  of  7500  amp.-turns  per  pole  when  an  e.m.f.  of  250 
volts  is  being  generated. 

Actually,  each  shunt  spool  is  wound  with  460  turns  of  No.  7 
B.  &  S.,  D.C.C.  wire.  The  field  current  is  therefore 

=  16.3  amp. 


The  shunt  coil  has  an  actual  length  of  6.25  in.     As  the  di- 
ameter of  No.  7  wire  is  0.16  in.,  including  insulation,  there  will  be 

Q-TQ  X  6.25  =  39  turns  per  layer  of  wire.     There  will  be  -^  = 

11.8  layers,  or  practically  12  layers,  giving  a  depth  of  winding 
of  0.16  X  12  =  1.92  in. 

The  mean  radius  of  the  coil,  allowing  for  spool  thickness,  is 
then 

Mean  radius  =  -  '• — ~ —  -  =  5.46  in. 

.*.  Mean  length  of  turn  =  2ir  X  5.46  =  34.35  in. 
Total  length  of  wire  on  each  shunt  spool  is 

-r^-  in.  X  460  =  1316  ft. 


Resistance  of  No.  7  wire  at  65°C.  =  0.586  ohm  per  1000  ft. 
.'.  Resistance  of  each  shunt  spool  is 

0.586  X  1.316  =  0.77  ohm. 
The  resistance  of  the  entire  shunt  field  is 

rf  =  0.77  X  12  =  9.24  ohm. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    63 

The  voltage  drop  on  the  shunt  field  is 

i/rf  =  16.3  X  9.24  =  151  volts. 
The  voltage  drop  in  the  shunt  field  rheostat  is 
erh.  =  250  -  151  =  99  volts. 

The  Series  Field  Winding.— Consider  two  cases:  (1)  the  genera- 
tor to  be  flat-compounded,  (that  is,  the  no-load  and  the  full- 
load  voltages  are  equal,  as  specified),  and  (2)  the  generator  to 
be  5  per  cent,  over-compounded. 

In  the  first  case,  it  is  evident  that  the  shunt  field  ampere- 
turns  will  remain  the  same  at  full-load  as  at  no-load  since  the 
same  voltage,  250,  is  impressed  on  the  shunt  circuit. 

But  by  Tables  III  and  IV,  it  is  seen  that  at  full-load  there  will 
be  required  10,427  —  7500  =  2927  additional  amp. -turns.  These 
must  evidently  be  supplied  by  the  series  field  m.m.f. 


FIG.  53. 


The  actual  winding  consists  of  2>^  turns  per  pole.  Each  turn 
is  made  up  of  5  strips  of  conductor  in  parallel,  each  strip  being 
3^  in.  wide  by  0.095  in.  thick.  The  accomplishment  of  half 
a  turn  is  illustrated  in  Fig.  53  which  represents  the  arrangement, 
in  plan,  of  the  series  field  winding. 

The  series  field  current  must  then  be 

2927A.27. 

l*  =  ocx       "  =  H70  amp. 
2.5  turns 

This  means  that  with  full-load  current  2000  -  1170  =  830 
amp.  must  be  diverted  from  the  series  turns  by  a  shunt  con- 
nected in  parallel  with  them.  This  shunt  is,  in  practice,  usually 
composed  of  German  silver  strips  whose  length  is  so  adjusted  by 
test  as  to  divert  exactly  the  required  amount  of  current. 

In  the  second  case,  the  full-load  voltage  with  5  per  cent,  over- 
compounding  is  1.05  X  250  =  262.5. 

To  obtain  this  voltage  requires  the  addition  of  11,700  —  7500  = 
4200  amp.-turns  to  the  no-load  ampere-turns.  This  additional 


64  ELECTRICAL  ENGINEERING 

excitation  is  not  all  supplied  by  the  series  field  m.m.f.,  however, 
since  the  shunt  field  current  is  affected  by  the  increased  terminal 
voltage.     The  shunt  field  m.m.f.  now  consists  of  1.05  X  7500  = 
7875  amp.-turns. 
Therefore,  the  series  field  m.m.f.  must  consist  of 

11,700  -  7875  =  3825  amp.-turns. 
The  current  in  the  series  winding  is  then 

it  =    o  g    =  1530  amp. 

a.O 

The  current  diverted  through  the  shunt  to  the  series  field  is 

2000  -  1530  =  470  amp. 
The  shunt  field  current  is 

if  =  17.1  amp. 

Consideration  of  the  saturation  curves  will  show  that  this  is 
nearly  the  limit  of  over-compounding  for  this  machine.  If 
full-load  voltage  of  275  were  desired,  it  would  be  necessary  to 
add  another  half  turn  to  each  series  coil. 

The  series  field  m.m.f.  has  been  made  to  compensate  for  the 
armature  reaction  and  the  ir  drop  (assumed  2%  per  cent.)  in 
the  armature.  So  far  as  the  field  design  is  concerned,  this  is 
satisfactory.  These  calculations  are,  however,  only  approxi- 
mate and  the  actual  values  should  now  be  determined  from  the 
known  data  of  the  machine. 

Armature  Resistance.  —  Being  multiple  wound,  there  are  12 
paths  in  parallel  in  the  armature.  Each  path  includes  72  con- 
ductors, or  36  turns.  The  length  of  a  turn  is  twice  the  gross 
length  of  the  armature  plus  the  end  connections.  The  end 
connections  for  one  turn  may  be  taken  as  9  X  diameter  per  pole 
of  the  armature  =  9  X  5.33  =  48  in. 

Length  of  one  turn  is  thus  2  X  9  in.  +  48  in.  =  66  in. 

f\f\    \/    Q  £\ 

Length  of  one  path  of  36  turns  =  -  -  =  198  ft. 

iZi 

Since  the  area  of  each  effective  conductor  section  is  0.0675 
sq.  in.,  its  resistance  is  found  to  be  0.142  ohm  per  1000  ft.  at 
65°C. 

Resistance  of  one  path  is  thus  0.142  X  0.198  =  0.02812  ohm. 
Resistance  of  12  paths  in  parallel  is 

" 


=  0.00234  ohm. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    65 

The  true  armature  resistance  will  be  somewhat  less  than 
this  owing  to  the  intermittent  short-circuiting  of  coils  by  the 
brushes,  and  its  average  value  may  be  taken  as 

ra  =  0.00226  ohm. 
Voltage  drop  in  the  armature  is 

ea  =  rala  =  0.00226  X  2017  =  4.55  volts. 

Brush  Resistance. — There  is  always  a  drop  in  voltage  at  the 
brushes  due  to  the  true  brush  resistance  and  also  to  the  re- 
sistance of  the  sliding  contact  between  brushes  and  commutator. 
This  combined  resistance  has  no  definite  value  which  may  be 
calculated,  but  it  is  found  by  experiment  that  the  drop  which 
it  causes  amounts  to  2  volts  when  the  current  density  in  the 
brushes  is  30  amp.  per  sq.  in.  or  more,  while  for  densities  less 
than  30,  the  drop  is  proportional  to  the  current  density.  30 
amp.  per  sq.  in.  is  about  the  usual  current  density  in  brushes. 
Drop  across  brushes  is  thus  6b  =  2  volts. 

Series  Field  Resistance. — Total  thickness  of  series  conductor  = 
0.095  in.  X  5  strips  =  0.475  -in.  Area  of  series  conductor  = 
0.475  X  3.125  =  1.485  sq.  in.  Mean  radius  of  series  turn, 
allowing  ^32  m-  insulation  between  turns,  is  found  to  be  5.12  in. 

mean  radius.  3  turns  +  mean  radius,  2  turns 
Mean  radius  =  -  — 2 — 

(4.5  +  0.475  +  0.0313  +  0.233)  +  (4.5  +  0.475  +  0.0156) 

2 

5.24  +  5        R1_. 
— 2 =  5.12  m. 

.*.  Mean  length  of  series  turn  =  2  X  5.12  X  TT  =  32.2  in. 

,,     ,       .        .    ,.  12  X  32.2  X  2.5 

Length  of  series  winding  =  —        — r~ —         -  =  80.5  ft.  approx. 

To  this  should  be  added  about  5  ft.  for  connections  between 
coils,  making  the  series  winding  85.5  ft.  long.  Resistance 
per  1000  ft.  of  series  conductor  is  found  to  be  0.00645  ohm  at 
65°C. 

Series  field  circuit  resistance  is  therefore 

ra  =  0.00645  X  0.0855  =  0.000552  ohm. 

As  it  was  found  that  only  1170  amp.  go  through  the  series 
field  coils  at  full-load,  the  voltage  drop  on  the  series  field  wind- 
ing is 

ea  =  r8ia  =  0.00055  X  1170  =  0.645  volt. 


66  ELECTRICAL  ENGINEERING 

Total  voltage  drop  in  the  machine  is  therefore 

ea  +  eb  +  e8  =  4.45  +  2  +  0.645  =  7.095  volts. 

or    '        =  0.0284,  or,  approximately,  2.5  per  cent,  as  assumed. 
/oU 

If  the  assumption  of  percentage  drop  is  not  considered  to  have 
been  sufficiently  close,  the  magnetic  calculations  should  be 
repeated  using  the  new  percentage  just  found. 

Commutator  and  Brushes. — The  size  of  the  commutator  is 
determined  chiefly  by  the  brush  requirements.  The  number  of 
commutator  segments  is  432,  that  is,  one  segment  to  each  effective 
turn  on  the  armature. 

The  brushes  rest  perpendicularly  on  the  commutator.  There 
are  12  studs  of  brushes,  each  stud  holding  10  brushes.  Each 
brush  has  a  cross-section  of  1.25  in.  X  0.75  in.,  giving  a  brush 
area  of  0.94  sq.  in.,  or  9.4  sq.  in.  per  stud. 

As  there  are  six  positive  and  six  negative  studs,  the  area 
of  the  positive  (or  negative)  brushes  is  6  X  9.4  =  56.3  sq.  in. 

Therefore  the  current  density  in  the  brushes  at  full-load  is 

2016.3 
,a  0     =  35.8  amp.  per  sq.  in. 

OD.o 

The  commutator  length  must  exceed  that  of  the  brushes  on 
the  stud,  that  is,  it  must  exceed  10  X  1.25  +  some  space  of 
separation  between  adjacent  brushes.  In  this  case  the  com- 
mutator length  is  17.5  in. 

The  commutator  diameter  is  influenced  by  the  peripheral 
speed.  Being  built  up  of  numerous  copper  segments  each 
separated  by  sheets  of  mica,  the  commutator  is  usually  mechan- 
ically weaker  than  any  other  revolving  part.  It  must  not  only 
be  protected  from  forces  which  would  cause  it  to  fly  apart,  but 
there  must  be  no  force  acting  upon  it  which  will  be  strong 
enough  to  cause  even  slight  warping  of  its  surface.  Good  com- 
mutation demands  smooth,  even  contact  between  the  segments 
and  the  brushes  at  all  times. 

On  the  other  hand,  too  small  a  diameter  results  in  very  narrow 
segments,  thin  and  wide  brushes  and  then,  in  turn,  a  longer 
commutator. 

The  commutator  diameter  for  this  machine  is  39  in.,  which  is 
approximately  60  per  cent,  of  the  armature  diameter.  From 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    67 

this  it  is  found  that  the  width  of  segment  plus  the  mica  in- 
sulation is 

7r39 


432 


=  0.284  in. 


0.75 


The  brushes  will  therefore  extend  over  Q  *       =  2.64  segments. 

Flux  Distribution  Around  the  Armature. — It  is  of  interest  at 
this  point  to  investigate  the  distribution  of  the  flux  around  the 
armature  periphery  on  account  of  its  bearing  on  the  commuta- 
tion and  also  in  order  to  be  able  to  determine  the  potential 
difference  between  any  two  adjacent  commutator  segments. 
This  is  best  accomplished  with  the  help  of  a  diagram  in  which  is 
shown  a  pair  of  poles  drawn  to  scale  in  relation  to  the  armature, 
developed  along  the  horizontal  line. 


FIG.  54. 

A  curve  abcde,  Fig.  54,  is  first  constructed  to  represent  the 
flux  distribution  around  360  electrical  space  degrees  of  the 
armature  periphery.  This  curve  is  based  on  the  assumption  of 
flux  density,  being  inversely  proportional  to  the  flux  path  in  the 
air.  Thus,  the  density  is  uniform  under  the  pole  and  is  so 
represented  by  the  line  ab.  To  determine  the  densities  between 
the  poles,  empirical  mean  flux  paths  to  the  teeth  are  drawn,  and 
the  flux  along  each  path  is  taken  as  inversely  proportional  to 
its  length.  The  curve  cde  will  obviously  be  the  reverse  of 
curve  abc. 


68  ELECTRICAL  ENGINEERING 

The  second  step  is  the  construction  of  a  curve  of  armature 
magnetomotive  force.  This  m.m.f .  will  act  in  the  direction  of  an 
axis  midway  between  the  poles  (assuming  brushes  to  be  set  on 
the  geometrical  neutral). 

Along  this  axis  the  m.m.f.  will  consist  of  all  the  armature 
ampere-turns  per  pole.  Acting  through  the  next  adjacent  teeth 
s,  s,  the  m.m.f.  will  be  diminished  by  the  amount  of  armature 
ampere-turns  included  between  these  teeth.  These  ampere-turns 
may  be  plotted,  tooth  by  tooth,  in  the  manner  thus  indicated, 
and  the  result  will  be  a  curve,  fgh,  in  the  form  of  successive  steps 
corresponding  to  the  armature  teeth.  To  construct  the  flux 
curve  of  the  armature  reaction  from  the  m.m.f.  curve,  reluctance 
of  the  air  paths  alone  need  be  considered.  To  be  sure,  the  rest 
of  the  flux  path,  especially  that  of  the  teeth,  would  have  some 
effect  on  the  accuracy  of  the  curves  so  obtained.  But  the  error 
would  not  be  great,  being  anywhere  from  2  per  cent,  to  8  per 
cent,  according  to  the  position  of  the  point  on  the  curve.  The 
flux  density  for  each  tooth  is  therefore  determined  from  the 
formula: 

3.19A.77. 
B  -        -z , 

where  I  is  the  length  of  the  path  in  air. 

1     This  is  plotted  as  curve,  ijk,  to  the  same  scale  as  the  curve  of 
the  field  flux  density — abode. 

The  actual  densities  along  the  periphery  will  vary  from  tooth 
to  slot,  and,  indeed,  this  variation  is  noticeable  on  many  oscillo- 
grams  of  alternator  voltage.  The  ripples  which  occur  in  the  flux 
wave  due  to  alternate  teeth  and  slots  would  exist  equally  with 
reference  to  the  field  flux,  armature  flux  and  resultant  flux. 
In  order  t6  avoid  confusion  the  ripples  have  not  been  shown  on 
the  armature  density  curve,  but  all  the  waves  are  plotted  as 
smooth  lines. 

A  study  of  the  resultant  wave  reveals  the  great  distortion  caused 
by  the  armature  current,  the  strengthening  of  the  flux  in  the 
pole  tips  at  A,  the  weakening  at  B,  and  the  shifting  of  the  neutral 
point,  c,  in  the  direction  of  rotation. 

The  student  may  well  discuss  fhe  effect  on  the  flux  density 
waves  of  giving  a  shift  of  the  brushes. 

Losses  and  Efficiency. — The  .efficiency  of  a  machine  is  given 
by  the  equation, 

output 
efficiency  =  ,  =  ---- 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    69 


The  full-load  output  of  the  generator  has  been  given 
P  =  El  -*-  1000  =  500  kw. 


as 


The  problem  of  the  efficiency  is  then  one  of  determining  the 
The  losses  of  a  generator  may  be  considered  under  three 
heads:  (1)  copper  losses,  due  to  heat  developed  by  the  currents  in 
the  windings;  (2)  core  losses,  due  to  hysteresis  and  eddy  cur- 
rents set  up  by  the  changes  of  magnetic  flux  in  the  iron  and,  to  a 
slight  extent,  in  the  copper  of  the  machine  and  (3)  friction  losses, 
including  that  of  the  bearings,  the  brushes  and  windage. 

Copper  Losses. — These  consist  of  I2r  loss  in  the  armature, 
the  shunt  field  circuit  including  the  rheostat,  the  series  field 
coils,  and  that  of  the  brushes  and  commutator. 

It  is  not  sufficient  to  ascertain  these  losses  for  full-load  only. 
The  quality  of  a  generator  is  displayed  by  its  performance  at  all 
reasonable  loads.  The  efficiency  will  in  this  case,  therefore, 
be  calculated  for  loads  from  zero  to  150  per  cent,  of  full-load. 

The  armature  copper  loss  is  7aVa,  where  ra  =  0.00226  ohm. 
Shunt  field  copper  loss  is  I/Ef,  where  E/  is  the  voltage  impressed 
on  the  field  circuit,  and  is  in  this  case  250  volts.  //  =  16.3 
amp. 

.'.  Shunt  field  copper  loss  =  16.3  X  250  =  4075  watts. 

The  series  field  loss  is  ISES,  where  I8  is  the  line  current  = 
Ia  —  If,  and  Ea  has  been  found  to  be  0.645  volts  at  full-load  and 
varies  directly  with  //  for  other  loads. 

TABULATION  OF  COPPER  LOSSES 


Per 

cent. 

0 

25 

50 

75 

100 

125 

150 

load 

la 

16 

516 

1,016 

1,516 

2,016 

2,516 

3,016 

/a2 

256 

266,000 

1,037,000 

2,300,000 

4,075,000 

6,330,000 

9,100,000 

/02r0 

0.58 

600 

2,340 

5,200 

9,200 

14,300 

20,550 

ItEf 

4,075 

4075 

4,075 

4,075 

4,075 

4,075 

4,075 

L 

0 

500 

1,000 

1,500 

2,000 

2,500 

3,000 

E, 

0 

0.161 

0.322 

0.484 

0.645 

0.806 

0.968 

I.E, 

0 

81 

323 

725 

1,290 

2,018 

2,905 

Eb 

0 

0.6 

1.2 

1.8 

2 

2 

2 

IaEb 

0 

310 

1,220 

2,730 

4,032 

5,032 

6,032 

Total 

loss 

4,075 

5,066 

7,958 

12,730 

18,597 

25,425 

33,562 

70  ELECTRICAL  ENGINEERING 

Brush  loss  =  IaEb,  where  Eb  is  the  voltage  drop  in  com- 
mutator and  brushes,  being  approximately  proportional  to  cur- 
rent density  in  the  brushes  up  to  a  density  of  30  amp.  per  sq. 
in.  and  being  2  volts  for  higher  current  densities. 

Core  Loss.  —  The  hysteresis  loss  is  principally  in  the  armature 
and  is  due  to  the  reversal  of  direction  of  the  flux  in  the  metal  as 
the  armature  spins  around.  The  amount  of  energy  expended 
in  reversals  of  the  magnetic  molecules  is  proportional  to  the 
frequency  and  approximately  proportional  to  (flux  density)1-6. 
Thus, 

Hysteresis  loss  =  kfB1-6. 

The  exponent  1.6  was  found  experimentally,  by  STEINMETZ; 
it  holds  with  sufficient  accuracy  for  the  usual  range  of  flux  densities 
obtained  in  electrical  machinery. 

In  direct-current  armatures  hysteresis  loss  usually  amounts  to 
about  2.8  watts  per  Ib.  at/  =  60  and  B  =  64,500.  Assuming  this 
value  as  standard,  the  armature  core  loss  and  teeth  loss  are  ex- 
pressed by  the  equation 

Wh  =  2.8  X  j4  X  (54  KQQ)  L8  X  wt.  of  core  or  teeth  in  Ib. 

For  both  core  and  teeth,  /  =  37.5. 

B,  in  core  =  43,600  at  250  volts,  full-load.     B,  corresponding 
to  average  amp.-turns  required  by  the  teeth  =  126,000. 
Weight  of  armature  core  =  vol.  X  wt.  of  1  cu.  in. 

=  0.28  X  6.05  X  7r(30J2  -  222)  =  2430  Ib. 

Weight   of  teeth  =  0.28  X  6.05  X  k(322  -  3O72)  - 

216  X  1.3  X  0.465] 
=  0.28  X  6.05  X  [258  -  130]  =  217  Ib. 

Substituting  these  values,  the  total  hysteresis  loss  in  teeth 
and  core  is 


=  1.75  [0.6761-*  X  2430  +  L951-6  X  217]  =  3340  watts. 

The  eddy  current  loss  is  due  to  the  heating  of  the  core  by  local 
or  eddy  currents  set  up  in  the  material  of  the  core  by  the  chang- 
ing flux  within  it. 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    71 

It  is  therefore  an  I2R  loss,  or  -~-,  where  E  is  the  e.m.f.  set  up, 
which  is  expressed  by  the  equation 


where  <j>  =  B  X  area  =  total  flux. 

From  this  it  may  be  seen  that  the  eddy  current  loss  may  be 
written 

We  =  k^B*. 

The  eddy  current  loss  may  be  reduced  as  much  as  desired  by 
making  the  laminations  of  the  armature  core  sufficiently  thin. 
A  satisfactory  value  for  this  loss  may  be  obtained  by  assuming 
it  equal  to  the  hysteresis  loss.  In  that  case,  We  =  3340  watts 
and  the  total  core  loss  is 

Wc  =  2  X  3340  =  6680  watts. 

Losses  in  pole  faces  and  copper  due  to  eddy  currents  are  here  too 
small  to  consider. 

There  will  also  be  slight  changes  in  the  values  of  the  core 
loss  as  the  load  changes,  due  to  variation  in  magnetic  densities, 
especially  in  the  teeth.  This  variation  is  also  slight,  however, 
and  will  be  neglected. 

Friction  Losses.  —  Loss  due  to  brush  friction  is  based  on  a 
coefficient  of  friction  of  0.3,  and  a  brush  pressure  of  1.2  Ib. 
per  sq.  in.  of  brush  surface.  From  this,  the  friction  per  sq. 
in.  is  0.3  X  1.2  =  0.36  Ib.  Surface  area  of  one  brush  =  1.25  X 
0.75  =  0.9375  sq.  in. 

Total  brush  friction  force  is  then, 

F  =  0.9375  X  10  X  12  X  0.36  =  40.5  Ib. 
Power  loss, 

h        w"  -  Hm  x  746  watts- 

where 

r  =  radius  of  commutator  in  ft.  =  1.625 
and 

n  =  speed  in  r.p.m.  =  375. 
Thus, 

_        2w  X  1.625  X  375  X  40.5  X  746 

Wb  =  33,000  =  35°°  WattS' 

1  This  is  the  equation  for  induced  e.m.f.  in  a  transformer.  It  holds  also 
in  this  case  as  will  appear  later  when  the  transformer  is  studied. 


72 


ELECTRICAL  ENGINEERING 


Bearing  friction  and  windage,  together,  make  a  complicated 
loss  to  determine  with  accuracy.  This  loss  is,  however,  one 
which  may  be  assumed  with  quite  sufficient  accuracy  from  the 
data  obtained  in  practice.  A  fair  assumption  to  make  for 
generators  of  this  type  is  1  per  cent,  of  the  rated  output  of  the 
generator.  In  this  case  then 

Wf  =  500,000  X  0.01  =  5000  watts. 

Summary  of  Losses,  Output  and  Efficiency. — The  combined 
losses  of  the  generator  for  different  per  cent,  loads  is  given  in 
Table  V. 


'60 


20 


50 


75 

%  Load 
FIG.  55. 


100 


125 


150 


TABLE  V 
The  efficiency  curve  is  shown  plotted  against  per  cent,  load  in  Fig.  55 


Per  cent,  load 

0 

25 

50 

75 

100 

125 

150 

Copper  loss  

4,076 

5066 

7  958 

12  730 

18  597 

25  425 

33  562 

Core  loss 

6680 

6  680 

6  680 

6  680 

6  680 

6  680 

6  680 

Friction  loss  .... 
Total  loss  

8,500 
19,256 

8,500 
20,246 

8,500 
23  138 

8,500 
27  910 

8,500 
33  777 

8,500 
40  605 

8,500 

48  742 

Output  
Input  

0 

19,256 

125,000 
145,246 

250,000 
273,138 

375,000 
402,910 

500,000 
533,777 

625,000 
665  605 

750,000 

798  742 

Efficiency  

0 

0  86 

0  915 

0  93 

0  936 

0  939 

0  939 

Temperature   Rise. — The   final   limit  to   the   output   of  the 
generator   is   the   permissible   temperature  rise.     The  effect  of 


DESIGN  OF  A  DIRECT-CURRENT  GENERATOR    73 

temperature  on  copper  is  to  increase  its  resistance  to  a  slight 
extent;  the  effect  on  iron  is  to  increase  its  permeability.  These 
effects  tend  to  offset  each  other  so  that,  as  tar  as  these  two 
materials  go,  it  would  be  permissible  to  attain  very  high 
temperatures.  . 

On  the  other  hand,  the  insulation  is  the  real  limiting  feature. 
Of  the  many  insulating  materials,  none  possesses  the  com- 
bination of  qualities  necessary  in  the  ideal  insulator  for  electrical 
machinery.  This  material  should  be  of  high  insulation  strength, 
strong  mechanically,  and  its  insulating  and  mechanical  qualities 
should  not  change  under  long-continued  heating.  Mica  is 
the  best  insulator  in  these  respects,  except  that  it  is  poor  from 
the  mechanical  standpoint.  Asbestos  is  useful  owing  to  its 
heat-resisting  qualities,  but  it  is  a  rather  poor  insulator  and  its 
mechanical  possibilities  are  limited.  Cotton  tapes  .and  varnishes 
do  not  withstand  the  high  temperatures. 

In  attempting  to  extend  the  limit  of  output  of  machines  of 
a  given  size  there  are  two  lines  along  which  lie  the  main  pos- 
sibilities of  success. 

Either  some  new  insulating  material,  more  satisfactory  than 
those  at  present  in  use,  may  be  discovered  or  invented,  or  im- 
provement in  ventilation  and  heat  radiation  may  be  accom- 
plished by  alteration  of  the  mechanical  design. 

Under  existing  conditions  a  temperature  rise  of  40°C.  above 
that  of  the  surrounding  air  is  quite  conservative.  The  tem- 
perature which  different  parts  of  a  machine  will  attain  is  hard 
to  predetermine  accurately  from  the  design.  Practical  studies 
have  afforded  certain  empirical  constants  which  permit  ap- 
proximate determinations  to  be  made,  but  in  any  case,  practical 
experience  will  greatly  assist  the  designer  in  his  attempts  to  keep 
close  to  the  limits. 

For  the  present  it  will  be  sufficient  to  determine  the  watts 
per  square  inch  of  surface  of  field  spools  and  armature.  For 
rotating  machinery  0.5  watt  per  sq.  in.  will  correspond  roughly 
to  a  temperature  rise  of  40°C. 

The  external  surface  of  a  field  spool,  only,  should  be  taken, 
and  the  same  applies  to  the  armature.  These  should,  of  course, 
be  calculated  separately. 

Problem  30. — In  the  machine  just  studied,  show  by  calculation,  as  indi- 
cated above,  that  the  temperature  rise  in  the  field  and  armature  coils  will 
not  be  excessive. 


CHAPTER  XI 

ELECTRICAL     CONSTANTS    OF     A     DIRECT-CURRENT 

GENERATOR  HAVING  COMMUTATING  POLES  AND 

COMPENSATING  WINDING 

As  a  typical  generator  of  this  more  complex  type  will  be  taken 
the  following: 

M.P.  6  -  1000  -  600  -  1200/1260  volts. 

The  generator  is  thus  5  per  cent,  over-compounded.  Being 
designed  for  comparatively  high  voltage,  commutation  becomes 
a  matter  of  special  importance. 

To  insure  proper  neutralization  of  the  armature  reaction,  there- 
fore, special  field  windings  are  supplied,  and  these  are  so  placed 
as  to  counteract  the  armature  m.m.f.  in  space  as  well  as  in 
amount.  That  is,  neutralization  is  accomplished  by  means  of  a 
compensating  winding  placed  in  the  pole  faces  symmetrically 
with  respect  to  the  armature  conductors  under  the  pole  arc,  and 
an  auxiliary  commutating  pole  inserted  between  the  main  poles, 
where  the  armature  magnetomotive  force  is  the  strongest,  and 
whose  duty  is  not  only  to  neutralize  this  magnetomotive  force 
about  the  neutral  point  in  which  the  brushes  are  placed,  but  to 
supply  a  flux  which  will  be  in  proper  direction  to  balance  the 
e.m.f.  of  self-induction  of  the  commutated  coil.  With  such  an 
arrangement  the  brushes  are  given  no  shift,  and,  consequently, 
the  armature  m.m.f.  is  entirely  cross-magnetizing. 

The  series  field  m.m.f.  proper  is  thus  relieved  of  every  duty  ex- 
cept those  of  compensating  for  IR  drop  in  the  armature  and  over- 
compounding.  The  circuit  diagram  of  this  machine  is  given  in 
Fig.  56. 

General  dimensions  and  specifications  are  as  follows: 

Armature  outside  diameter,  48  in. 

Armature  inside  diameter,  28  in. 

Armature  gross  length  diameter  15.5  in. 

Armature  effective  diameter,  11.7  in. 

Armature  ventilating  ducts,  4%  in.  wide. 

2%  in.  wide. 
74 


ELECTRICAL  CONSTANTS 


75 


Slots,  number  and  dimensions,  144;  0.44  in.  X  1.53  in. 
Effective  conductors  per  slot,  6. 

Effective  armature  conductor  section,  0.55  in.  X  0.09  in. 
Armature  winding,  multiple  drum. 


Compensating    Commutating 
Series          Winding  Poles 


Rheostat 


Compensating  Winding1 


FIG.  56. 

Yoke  section,  rounded,  17  in.  X  6.5  in. 
Main  pole  core  section,  14.5  in.  X  14.5  in. 
Main  pole  core  length,  including  pole  shoe,  14  in. 
Main  pole  core  length,  allowed  for  field  spool,  13  in. 
Main  pole  arc,  17.5  in. 


76 


ELECTRICAL  ENGINEERING 


Commutating  pole  section,  13.5  in.  X  2.25  in. 

Commutating  pole  length,  14  in. 

Main  air  gap  length,  0.3125  in. 

Air  gap  under  Commutating  pole,  0.5  in. 

Shunt  field  winding;  2256  turns  per  spool  of  No.  15  B.  &  S. 
triple  cotton-covered  wire. 

Series  field  winding;  3  turns  per  spool.  Each  conductor  built 
up  of  4  strips  giving  total  section,  1.5  in.  X  0.35  in. 

Commutating  pole  winding;  5.5  turns  of  copper  ribbon  12  in. 
wide  X  0.05  in.  thick. 


1400 


1000 


•800 


600 


400 


200  400  600  800 

Ampere-Turns 

FIG.  57. 


1000 


1200 


Compensating  (pole  face)  winding  consists  of  16  conductors  per 
pole  contained  in  8  holes  in  the  pole  face.  Each  hole  has  2 
conductors,  one,  a  tube,  the  other  a  rod  within  the  tube.  Tube 
outside  diameter,  1%  in.,  inside  diameter  2%2  in. 

Rod  diameter,  %  in. 

Commutator  diameter,  30  in. 

,  Commutator  length,  14  in. 

Commutator  segments,  432. 

Segment  width,  0.219  in. 


ELECTRICAL  CONSTANTS  77 

Brushes  per  stud,  7. 

Brush  section  dimensions,  1.25  in.  X  0.875  in. 

The  armature  flux  at  no-load  is  readily  found  to  be  13.9 
megalines  per  pole. 

The  no-load  saturation  curve  is  given  in  Fig.  57,  having  been 
determined  in  exactly  the  same  manner  as  that  of  the  previous 
machine,  given  in  Fig.  52. 

This  curve  shows  that  7600  amp.-turns  are  required  to  give 
normal  voltage  at  no-load.  At  full-load,  the  shunt  field  m.m.f. 

1260 
will  supply  J2QO  X  760°  =  798°  amP--turns- 

Assuming  2  per  cent,  voltage  drop  in  armature  and  brushes,  the 
total  e.m.f.  which  must  be  generated  is  1.02  X  1260  =  1285 
volts.  From  the  saturation  curve,  this  voltage  requires  8750 
amp.-turns.  Therefore  the  series  field  m.m.f.  must  supply 
8750-7980  =  770  net  amp.-turns  per  pole. 

To  supply  these,  however,  account  must  be  taken  of  the  un- 
fortunate situation  of  the  series  field  winding  with  respect  to 
magnetic  leakage.  Being  placed  close  to  the  yoke,  the  leakage 
factor  should  probably  be  1.50  instead  of  1.25  as  used  for  the  shunt 
field  calculation.  This  factor  could,  of  course,  be  calculated, 
but  it  is  hardly  desirable  to  introduce  such  a  refinement  when 
the  means  of  adjustment  of  the  series  field  current  render  a  reason- 
able assumption  entirely  satisfactory.  On  the  basis  of  a  leakage 

1  50 
factor  of  1.50,  the  series  amp.-turns  are  T^  X  770  =  924. 

924 
The  series  field  current  is  I8  =  -$-  =  308  amp. 

o 

Current  diverted  around  the  series  field  is 

Id  =  793  -  308  =  485  amp. 

The  entire  load  current  of  793  amp.  passes  through  the  9  turns 
per  pole  of  the  compensating  winding,  and  the  5J^  turns  of  each 
commutating  pole. 

SATURATION  CURVE  CALCULATION 

No-load.     E  =  1200. 

_  E  x  1Q8 

*"   =        4/< 
where 

600       6 
*  =  60    X  2  =  30' 


78 


ELECTRICAL  ENGINEERING 
6  X  144 


t  = 


x  2    =  72  turns  per  pole. 
1  200  ^  1 


=  13>880>000  =  flux  in  teeth 


••     «       307  X  72  X  4 
and  gap  at  no-load.     The  flux  in  the  pole  core  is 

fa  =  1.25  X  13,880,000  =  17,340,000, 

where  1.25  is  the  leakage  factor.     It  is  fairly  large  in  this  case,  as  is  usual 
when  interpoles  are  present. 


Tooth  width  at  face  = 
Teeth  per  pole  = 

Teeth  width  (face) 
Teeth  width  (base) 

Teeth  area  (face) 
Teeth  area  (base) 

Gap  area 
Arm.  core  area 


X48 
144 


-  0.44  =  1.05  -  0.44  =  0.61  in. 


pole  pitch 

10.85  in. 
X  10.85  =  10.15  in. 


24  X     X  1.08 
25.5 


=  17.75. 


17.75  X  0.61  in. 
44.92 


48 

10.85  X  11.7  =  127  sq.  in. 
10.15  X  11.7  =  118.8  sq.  in. 
3  X  17.5  X  14.5  +  127 


=  222  sq.  in. 


44.92  -  28 


X  11.7  =  99  sq.  in. 


Pole  core  area  =  14.5  X  14.5  =  210  sq.  in. 

Yoke  area  =  17.5  X  6.5  X  0.95  =  108  sq.  in. 

No-load.    E  =  1200 


Part 

Mate- 
rial 

Flux 

Area 

B 

AT  fin. 

Length 

AT 

Teeth  (face)  
Teeth  (base)  .... 
Gap.. 

13.88 
13.88 
13.88 

127 

118.8 
222 

109,200 
116,800 
62,500 

93  \ 

iso/136 

19,600 

1.53 
0.3125 

208 
6,125 

Arm  

Sheet 

Pole  

iron  .  . 
Steel 

6.94 
17.34 

99 
210 

70,000 
85,200 

7 
33.5 

9.5 
14 

67 
470 

Yoke  

Steel 

8  67 

108 

80,300 

30 

24.3 

730 

Total 

7,600 

E  =  600 


Teeth  (face)  

54,600 

3.9  1    . 

Teeth  (base)  .  .  . 

58,400 

4.4  I4'15 

6.35 

Gao 

3,063 

Arm  .         .    . 

35,000 

2.4 

22.8 

Pole 

41,250 

9 

126 

Yoke 

40,150 

8.8 

215 

Total 

3,433 

ELECTRICAL  CONSTANTS 

E  =  900 


79 


Teeth  (face)  .... 

82,000 

11  Ol 

Teeth  (base)   .  .  . 

87,500 

14  5/  12'75 



19.5 

Gap.. 

4,594 

Arm                   .  . 

52,500 

3  77 

35  8 

Pole 

61  900 

15  9 

222  5 

Yoke  

60,200 

15.1 

367 

Total 

5  239 

E  =  1400 


Teeth  (face). 

127,400 

420  1 

Teeth  (base) 

136,100 

1,264  f842 

1,290 

Gap. 

7,150 

Arm 

81,600 

10.8 

102.7 

Pole 

96,200 

79 

1,107 

Yoke            .... 

93,600 

63.8 

1,660 

Total 

11,310 

FIG.  58. 

Calculations  of  armature,  shunt  and  series  field  windings,  as 
well  as  brush  losses  and  friction  loss  are  made  in  exactly  the 
same  manner  as  in  the  preceding  example.  The  difference  in 
location  of  the  shunt  and  series  windings  is  given  in  Fig.  58. 
The  division  of  the  shunt  into  two  coils  per  pole  is  made  to 


80         ELECTRICAL  ENGINEERING 

allow  the  necessary  room  for  end-connections  of  the  compensating 
winding. 

The  calculation  of  the  commutating  pole  winding  is  likewise 
a  matter  of  applying  the  old  principle. 

The  conductor  itself  is  of  extreme  dimensions,  being  a  band 
of  sheet  copper  1  ft.  in  width. 

For  the  compensating  winding  the  mean  length  of  1  turn 
is  found  to  be  2  X  (length  of  pole  parallel  to  shaft  +  4  in. 
(extension))  +  2  X  mean  span  between  poles,  =  2  X  (14.5  + 
4)  +  2  X  19  in.  =  75  in. 

Total  length  of  winding  =  -^  X  8  X  6  =  300  ft. 

Area  of  conductor  section  =  0.442  sq.  in. 

Resistance  of  winding  =  p-  =  -TQJ-  X  TTTIo  =  0-0054  ohm. 

Voltage  drop  in  winding  =  793  X  0.0054  =  4.28  volts. 
Loss  in  winding  =  4.28  X  793  =  3400  watts. 
Voltage  drops  and  losses  at  full-load  in  other  parts  of  the 
generator  are  as  follows: 

Voltage  drop  Loss 

Armature  ...........    ........  .....  14.75  11,700 

Shunt  field  ........................  (1,008) 

including  rheostat  ................  ......  4,500 

Series  field  ........................  0.636  505 

Compensating  winding  ..............  4  .  28  3,400 

Commutating  field  .................  1.19  945 

Brushes  (72#)  .....................  2  1,590 

Brushes  (friction)  ..................  ......  1,760 

Hysteresis  loss  .....................  .......  7,220 

Eddy  current  ......................  7,220 

Friction  and  windage  ...............  ......  10,000 

Total  voltage  drop  .............  22  .  856 

Per  cent,  voltage  drop  ......  ....  1  .  82 

Total  energy  loss  ....  r  .........  48,840 

output  1,000,000 

Efficiency  =  =  1,048,840  = 


Efficiencies  for  all  loads  are  as  follows: 

Per  cent,  load    0        25        50          75         100         125         150 
Percent,  eff.      0    88.5     93.5     94.85      95.3      95.5       95.5 

Fig.  59  shows  the  efficiency  curve. 


ELECTRICAL  CONSTANTS 


81 


60 


40 


25  50 


75  100 

%  Load 

FIG.  59. 


125  150 


O  Q  O  Ok 


/O  000 


FIG.  60. 


82  ELECTRICAL  ENGINEERING 

EFFECT   OF  COMPENSATING  WINDING  AND 
COMMUTATING  POLES 

To  study  the  effect  of  these  windings  in  neutralizing  armature 
reaction,  it  is  best  to  construct  a  curve  of  magnetomotive  forces 
showing  their  distribution  along  the  armature  periphery.  From 
this  and  the  curve  of  field  flux  density  the  resultant  flux  density 
along  the  periphery  is  obtained.  Such  curves  are  given  in  Fig. 
60.  The  armature  ampere-turns  and  field  flux  density  in  the  gap 
are  plotted  to  separate  scales  as  was  done  in  Fig.  54.  The  corn- 
mutating  pole  and  compensating  winding  ampere-turns  are  like- 
wise plotted,  but  their  direction  is,  of  course,  opposite  to  that  of  the 
armature  m.m.f.  The  resultant  m.m.f.  of  these  three  is  given 
by  the  heavy  irregular  line.  The  average  of  this  resultant  m.m.f. 
is  seen  to  be  very  nearly  zero,  showing  the  effective  compensation 
of  the  armature  reaction.  It  is  also  observable  that  the  commu- 
tating  pole  m.m.f.  is  made  sufficiently  strong  to  overbalance  con- 
siderably the  armature  m.m.f.  in  the  neutral  axis,  thus  creating 
a  resultant  flux  oppositely  directed  to  the  armature  m.m.f. 

The  maximum  armature  m.m.f.  which  acts  along  the  commu- 
tating  pole  axis  is  9564  amp.-turns.  Opposing  this  is  the  m.m.f. 
of  the  compensating  winding  which  is  6344  amp.-turns,  and  the 
m.m.f.  of  the  commutating  pole  which  is  4360  amp.-turns.  Thus 
the  resultant  amp.-turns  amount  to  (6344  +  4360)  —  9564  = 
1140.  When  the  armature  is  in  the  less  advantageous  position 
(that  is,  with  a  slot  in  the  commutating  pole  axis),  the  resultant 
amp.-turns  are  10,704  -  (11.5  X  797)  =  1554. 

The  average  resultant  amp.-turns  along  the  commutating  pole 
axis  are  therefore  1350. 

These  ampere-turns,  acting  through  a  gap  of  %  m-  produce  a 
flux  density  of 

1350 
B  =  0.313  X  0.5  =  862°  lines  per  Sq"  in' 

Commutation. — If  the  field  in  the  neutral  axis  were  completely 
neutralized,  commutation  would  still  be  poor  due  to  the  reversal 
of  the  current  in  the  conductors  during  the  period  of  commuta- 
tion. Therefore,  to  balance  the  e.m.f.  induced  in  the  short- 
circuited  coil  under  the  brush,  an  approximately  equal  e.m.f.  is 
created  in  the  opposite  direction  in  this  coil  by  causing  it  to  cut 
through  the  flux  due  to  the  commutating  pole. 

Exact  neutralization  of  the  induced  e.m.f.  in  the  short-circuited 


ELECTRICAL  CONSTANTS  83 

coil  is  practically  impossible  by  this  means.  Current  in  the 
conductors  does  not  vary  logarithmically  as  in  an  ordinary  circuit 
when  the  impressed  e.m.f.  is  removed.  If  it  did,  the  fundamental 
equation,  (15), 

e  =  ir  +  L  ^,  where  r  and  L  are   approximately   constant, 

would  apply  for  the  induced  e.m.f.  of  the  coil.  But  in  this  case, 
r  is  by  no  means  constant  due  to  the  varying  brush  surface  on  the 
commutator  segments.  The  value  of  r  is  therefore  some  func- 
tion of  the  time.  Putting  r  =  /  (Q,  and  considering  the  varia- 
bility of  L,  due  to  change  of  permeability  in  the  iron  part  of  the 
flux  path,  the  induced  e.m.f.  would  be  expressed  by  the  equation 

e  =  »7(0  +  I  (Li)- 

To  solve  this  equation  to  a  satisfactory  degree  of  approxima- 
tion, certain  assumptions  may  be  made.  First,  let  it  be  assumed 
that  the  current  dies  down  in  the  coil 
as  a  sine  wave  (Fig.  61).  The  in- 
duced e.m.f.  would  then  be  maximum 
when  the  coil  axis  passed  through  the 
center  of  the  brush. 

If  this  maximum  value   were  de- 
termined, it  could  be  made  equal  to  F 

the  e.m.f.  produced  by  rotation  of  the 

coil  through  the  field  set  up  by  the  commutating  pole.  Other 
values  than  the  maximum  could  be  left  to  care  for  themselves, 
being  of  secondary  importance.  The  maximum  value  of  the 
e.m.f.  of  self-induction  is 


where  I  is  the  current  in  the  armature  conductor  at  the  moment 
when  commutation  begins,  and,  in  this  case,  is  133  amp.  and  X  is 
the  reactance  of  the  coil. 

The  second   assumption  is  that  L,  the   coil  inductance,  is 
constant.     Hence 

X  =  27r/cL, 

where  fc  =  frequency  of  current  during  commutation. 

fc  =  ^TTT  where  Tc  is  the  time  of  cummutation,  since  this  time 

"I  c 

evidently  corresponds  to  one-half  wave  length.     The  time   of 


84  ELECTRICAL  ENGINEERING 

commutation  is  that  time  taken  by  the  commutator  to  move  a 
distance  equal  to  the  thickness  of  a  brush.  In  the  machine  under 
consideration  each  brush  covers  four  segments. 


/.  Tc   =  -±-  X 


segments  covered  by  brush 


r.p.s.  total  segments 

1          4 

=  TO  X  432  =  0'000952  sec'» 
and 


2  X  0.000925  =  54°  cycles  per  sec" 


In  calculating  the  coil  inductance,  L,  it  is  not  sufficient  to  consider 
only  the  interlinkage  of  each  coil  with  the  flux  which  it  produces. 
Mutual  induction  is  also  present,  the  value  of  L  desired  being 
therefore  not  strictly  the  self-inductance,  but  including  that  due 
to  the  interlinkage  of  the  flux  produced  by  the  current  in  all 
6  turns  with  each  single  turn.  In  this  machine  the  conductors 
in  each  slot  are  all  in  parallel;  thus  N  is  1  turn,  composed  of  2 
conductors.  It  should  be  noted  that  of  the  2  conductors  com- 
posing any  turn,  one  of  them  lies  in  the  lower  half  of  its  slot,  while 
the  other  lies  in  the  upper  half  of  its  slot.  The  interlinkage  of 
each  conductor  with  the  total  flux  will  not  be  the  same  in  the  2 
cases. 

However,  by  considering  the  total  flux  as  due  to  the  67  amp.- 
turns  of  a  slot  acting  through  an  effective  magnetic  conductance, 
G,  and  surrounding  each  of  the  6  conductors,  the  inductance 
thus  calculated  will  be  correct,  provided  the  proper  value  of  G 
is  determined.  Thus, 

<t>  =  67  X  G, 

where  <£  is  the  equivalent  flux  surrounding  all  conductors  in  1 
slot.  (The  inductance  due  to  end-connections  must  also  be 
ascertained,  as  is  done  later.) 

The  magnetic  conductance  per  centimeter  effective  length  of 
the  armature  is  calculated  by  means  of  the  general  formula, 

area 

x 


Considering  the  magnetic  circuit  (Fig.  62),  it  is  seen  to  consist 
of  3  parallel  paths  in  air,  namely:  that  of  section  A  and  length 
B  through  the  conductors,  that  of  section  C  and  length  B  above 


ELECTRICAL  CONSTANTS 


85 


the  conductors,  and  that  of  section  F  and  length  D,  from  the  top 
of  1  tooth  to  the  top  of  the  other.  The  common  path  through 
the  iron  may  be  neglected  as  offering  comparatively  little 
resistance. 

To  find  the  effective  magnetic 
conductance  per  centimeter 
length  across  the  section  A ,  con- 
sider an  elementary  section,  dx, 
at  distance,  x,  from  the  bottom 
of  the  conductors.  The  con- 
ductance across  this  section  is  — ' 
QAirdx 


B 


FIG.  62. 


The  amp.  -turns  acting  in  this  conductance  are  —  T-,  where  /  is 

** 


in  amperes. 

Therefore  the  flux  set  up  through  dx  is 


d<f> 


QAirdx 
~B~ 


X 


2.47T/  x  dx 
AB      ' 


This  flux  interlinks  with  only  -r-  conductors. 

A. 


Nd+  =  -L-AJ8— ' 


14.47T/  r\ , 

z2cfo 


Thus, 
and 


14.47T/A 

SB 

is  the  interlinkages  of  the  flux  with  all  6  conductors. 

Thus,  if  the  flux  is  considered  to  be  due  to  67  amp.-turns 
acting  through  conductance,  g,  and  this  interlinks  with  each 
conductor,  the  total  number  of  interlinkages  is 

14.4rr7A 


whence, 


14.47T/A 


~  3B/X36  ~      3B 
The  other  two  paths  are  entirely  outside  of  the  conductors, 


86  ELECTRICAL  ENGINEERING 

and  hence  are  acted  on  by  all  of  the  ampere-turns  in  the  slot. 
These  conductances  are  then,  respectively, 

0.47T  ^  and  0.47r  y:- 
Hi  U 

The  total  effective  conductance  is  then 

A    .   C    .  F 


per  cm.  length  of  armature,  and  the  flux  per  slot  is 
6  =  ING  X  2.54? 


where  I  is  the  effective  length  of  the  armature,  in  inches.     For 
both  slots  and  1  complete  turn  the  inductance  is 

<t>N       2  X  6X3.2  X  11.7in.rA        C       F 


a  ~  1087  108 

Substituting  numerical  values  for  the  slot  and  tooth  dimensions, 

A  =  1.233  B  =  0.44  C  =  0.213 

D  =  1.047  E  =  0.51  F  =  0.607, 

this  inductance  becomes 

L,  =  449  [0.934  +  0.417  +  0.58]  X  10~8  =  867  X  10~8  henrys. 
To  this  must  be  added  the  inductance  of  the  end-connections. 
The  flux  produced  by  the  end-connections  per  ampere-turn  per 
inch  of  coil  length  may  be  taken  as  one-twentieth  of  that  in  the 
slot. 

It  is,  therefore,  ^  X  3.2  X  1.931  =  0.309  lines. 
zo 

The  coil  divides  as  it  passes  out  from  the  slot,  so  that  only 
3  conductors     3   conductors   are  grouped    together.      There- 
fore   there    are  37   amp.-turns   producing   flux 
around  each  conductor.     If  the  length  of  the 
end-connections  for  1  turn  is  assumed  as  8  X 
diameter  per  pole,  =  8  X  8  =  64  in.,  the  flux 
surrounding   each  turn  is   </>  =  0.309  X  37  X 
FIG.  63.  64  =  59.47  lines.     The  inductance  is  then 

T  <t>N          59.4  X  1  X  7 

Le  =  TxW=        10*  X  7        =  59.4  X  10-  henrys. 

The  total  inductance  is  thus 

L  =  L,  +  Le  =  (867  +  59.4)  X  10~8  =  926.4  X  10~8  henrys. 


ELECTRICAL  CONSTANTS  87 

The  reactance  of  the  short-circuited  coil  is 

X  =  2irfcL  =  6.28  X  540  X  926.4  X  10~8  =  0.0314  ohm, 
and  the  maximum  e.m.f.  of  self-induction  is 

Em  =  IX  =  133  X  0.0314  =  4.17  volts. 

To  overcome  this  e.m.f.  the  short-circuited  coil  is  made 
to  rotate  in  the  field  of  the  commutating  pole.  This  field 
has  been  found  to  have  an  average  density  around  the  neutral 
axis  of  about  8620  lines  per  sq.  in. 

In  this  case,  the  commutated  coil  has  only  one  turn.  Thus 
e.m.fs.  are  generated  in  one  conductor  under  a  " north"  com- 
mutating pole,  and  in  the  other  conductor  under  a  " south" 
commutating  pole.  These  e.m.fs.  are  similar,  and  together 
make  up  the  total  e.m.f.  generated  in  the  coil  by  rotation  in  the 
commutating  field.  The  maximum  value  of  this  induced  e.m.f. 
corresponds  to  the  rate  of  cutting  the  flux  in  the  center  under  the 
commutating  pole. 

Consider  a  small  distance,  dx,  Fig.  64,  at  this 
point.  The  flux  through  the  area  of  width,  dx, 
and  average  length,  11.7  in.,  of  the  iron  in  field 
and  armature  is  FIG.  64. 

d<f>  =  8620  X  11.7  dx  =  101,000«te. 

The  speed  of  conductors  at  the  armature  periphery  is  irD  X 
r.p.s.  =  TT  X  48  X  10  =  4807T  in.  per  sec. 

The  time  required  for  a  conductor  to  go  the  distance  dx,  is 

dT-  — 
dl       480ir 


, 

•  •   *  induced    =    10^  -  J^_         =    L53   V°ltS 

J  4807T 

per  conductor,  or  3.06  volts  per  coil. 

This  voltage  opposes  that  due  to  self-induction,  leaving 
as  a  resultant, 

4.17  -  3.06  =  1.11  volts 

acting  in  the  circuit. 

Since  experience  has  taught  that  two  volts  potential  difference 
can  be  taken  care  of  by  the  resistance  of  the  carbon  brush  no 
difficulties  from  sparking  need  be  anticipated. 


CHAPTER  XII 

DIRECT-CURRENT  GENERATORS  IN  PARALLEL 
AND  SERIES 

Shunt  generators  operate  in  parallel  without  the  slightest  diffi- 
culty. Generator  No.  1  is  first  started  and  thrown  on  the  line. 
Generator  No.  2  is  then  brought  up  to  about  normal  speed,  the 
voltage  is  adjusted  and  the  line  switch  is  closed.  Since  genera- 
tor and  line  voltage  are  the  same,  no-load  is  taken  by  generator 
No.  2.  By  adjusting  the  field  excitation  of  No.  2  the  generator 
takes  the  desired  share  of  the  load.  As  its  load  increases  its 
engine  slows  down,  the  governor  opens  and  the  speed  is  restored 
to  normal. 

Series  generators  do  not  operate  naturally  in  parallel.  Assume, 
for  example,  that  two  series  generators  are  in  parallel,  each 
taking  its  share  of  the  total  load.  Suppose  then  that  for  some 

. .    reason  the  voltage  of  No.  2  (Fig.  65) 

_  J j  becomes  slightly  reduced.     Its  share  of 

the  load  will  fall  off  proportionately 
and,  with  this,  its  field  excitation. 
Falling  off  of  the  field  excitation 
further  reduces  the  voltage  and,  con- 
sequently, the  load,  the  excitation, 

and  so  on.  The  current  is  reduced  to  zero,  then  reversed  in 
direction  in  both  the  field  and  the  armature  coils.  The  rotation 
of  No.  2  remains  the  same,  but  the  machine  now  acts  as  a  series 
motor  driving  its  engine.  In  practice,  the  rush  of  current  dur- 
ing this  period  when  the  counter  e.m.f.  of  generator  No.  2  has 
been  destroyed  is  so  great  that  the  circuit  is  opened  by  its  fuses 
or  circuit  breakers.  Series  generators  are  not  in  common  use, 
but  this  principle  of  instability  in  parallel  operation  applies 
equally  to  compound  generators  through  their  series  field 
windings. 

With  shunt  generators  there  is  no  such  instability.  If  the 
voltage  of  No.  2  falls  off,  its  current  likewise  is  reduced.  But  the 
effect  of  reduced  current  is  to  lessen  the  armature  reaction,  thus 

88 


DIRECT -C  URRENT  GENERA  TORS  89 

bringing  up  the  voltage.     The  shunt  field  current  is  not  affected 
since  it  is  derived  from  the  bus  bars. 

Series  generators  and,  more  particularly,  compound  generators 
may  be  made  stable  in  parallel  operation  by  the  use  of  an 
' 'equalizer  bus."  This  consists  of  a  very  heavy  copper  connection 
situated,  as  shown  in  Fig.  66,  between  the  inner  terminals  of  the 
series  field  circuits  of  the  two  (or  more)  generators.  If,  now, 
the  voltage  of  No.  2  becomes  reduced  to  a  slight  extent,  current 
will  flow  from  the  +  brush 
of  No.  1  through  the 
equalizer  and  into  the  series 
field  coils  of  No.  2,  main- 
taining the  strength  of  the 
field  of  the  latter. 

If    the    two   generators, 
in    normal    operation,    do  F 

not  divide  the  load  prop- 
erly in  the  proportion  of  their  respective  ratings,  this  may 
be .  corrected  by  inserting  resistance  in  the  series  field  circuit 
of  that  generator  which  takes  too  much  of  the  load.  The 
effect  of  the  equalizer  is  to  put  the  series  field  coils  always  in 
parallel.  The  voltage  across  these  coils  is  therefore  the  drop 
between  the  positive  brushes  and  the  positive  bus.  The  re- 
sistance of  the  equalizer  is  so  low  that  its  drop  is  negligible,  so 
that  the  drop  across  all  the  series  field  coils  is  the  same.  Putting 
a  shunt  or  diverter  around  one  of  the  series  field  coils  has  no 
effect  on  the  distribution  of  the  load  on  any  particular  generator, 
as  it  affects  all  the  series  field  currents  alike,  the  proportions 
remaining  the  same. 

Direct-current  Generators  in  Series. — No  inherent  difficulty 
is  encountered  in  connecting  direct-current  generators  in  series. 
Owing  to  the  limited  possibilities  of  constructing  commutators 
that  will  permit  the  generation  of  very  high  voltages,  where 
these  are  required  in  direct-current  machines  recourse  is  usually 
had  to  series  connection. 

In  electric  railway  work  it  is  the  general  rule  to  employ  both 
series  and  parallel  connection  of  the  motors  to  give  flexibility 
in  speed  control. 

The  Three -wire  System. — Two  generators  in  series  afford  the 
simplest  means  of  obtaining  the  three-wire  system.  This  system, 
invented  by  EDISON,  was  devised  to  enable  the  use  of  large 


90 


ELECTRICAL  ENGINEERING 


E 


numbers  of  low  voltage  incandescent  lamps  without,  at  the  same 
time,  entailing  the  use  of  a  prohibitive  amount  of  copper  in  the 
distribution  system.  As  seen  in  Fig.  67,  the  voltage  of  the 
system  is  2Et  while  that  across  any  ele- 
ment of  the  system  is  only  E. 

There  are  other  ways  by  which  power 
may  be  supplied  to  such  a  system.  Thus, 
the  source  of  power  may  be  a  single  gen- 
erator of  voltage,  2E,  across  whose  termi- 
nals may  be  connected  either  a  storage 
battery,  as  in  Fig.  68,  or  two  small  gener- 
ators mounted  on  the  same  shaft,  called  a 
FIG.  67.  balancer,  and  shown  in  Fig.  69.  In  either 

case  the  necessary  condition  is  to  have  available  some  connec- 
tion point  the  potential  of  which  is  intermediate  between  those 
of  the  outer  wires.  The  amount  of  current  actually  flowing  in 


FIG.  68. 


FIG.  69. 


either  the  battery  or  the  balancer  set  is  small  in  case  the  two 
sides  of  the  load  are  reasonably  well  balanced. 

Another  scheme  consists  in  the  use  of  the  three-wire  generator. 
This  is  illustrated  in  Figs.  70  and  71.  Fig.  70  shows  a  bi-polar 
machine  constructed  by  reversing  the  windings  on  two  adjacent 


FIG.  70. 


FIG.  71. 


poles  of  a  four-pole  generator.  The  potentials  of  the  two 
brushes  on  the  horizontal  axis  are  the  same  and  are  midway 
between  the  potentials  of  the  two  other  brushes.  The  object  of 
making  the  machine  bi-polar  is  to  give  an  intermediate  inactive 


DIRECT-CURRENT  GENERATORS  91 

belt  along  the  commutator  on  which  a  brush  may  be  placed 
without  causing  disruptive  sparking.  A  better  scheme  is  'that 
of  DOBROWOLSKY  shown  in  Fig.  71.  The  armature  is  tapped  at 
two  opposite  points  which  are  connected,  through  slip  rings,  to  a 
" choke"  coil,  which  is  simply  an  induction  coil.  This  coil  is 
wound  upon  a  laminated  iron  core,  and  therefore  is  of  high  in- 
ductance. The  e.m.f.  impressed  upon  it  is  evidently  alternating, 
and  therefore  very  little  alternating  current  can  flow  through 
the  coil. 

The  middle  point  of  the  coil  must  always  be  at  a  potential 
midway  between  those  of  the  brushes.  It  may  therefore  be 
connected  to  the  middle  wire  of  the  system. 

The  disadvantage  of  using  a  battery  is  that  some  cells  may  be 
called  on  to  supply  more  energy  than  others.  It  then  becomes 
difficult  to  keep  the  battery  uniformly  charged,  and  deteriora- 
tion results. 

No  such  difficulty  occurs  with  the  use  of  balancers.  They 
may  be  small,  inexpensive  machines,  which  when  running  idle 
take  only  a  small  current. 

As  an  example  of  the  use  of  balancers  and  the  economy  of  the 
three-wire  system,  consider  the  circuit  illustrated  in  Fig.  72. 


20 


FIG.  72. 

The  load  consists  of  40  amp.  on  the  upper  branch  and  30  amp. 
on  the  lower.  The  system  is  therefore  unbalanced.  Currents 
and  directions  of  flow  are  indicated  for  each  portion  of  the  circuit. 
The  current  in  the  middle  or  neutral  wire  varies,  being  10  amp. 
in  some  sections  and  0  in  others.  Let  it  be  assumed  that  the 
current  required  to  run  the  balancer  set  is  1  amp.  which  would 
be  indicated,  if  shown  in  the  figure,  by  an  arrow  pointing  down- 
ward in  the  balancer  set.  The  current  returning  to  the  balancer 
over  the  middle  wire  is  10  amp.  This  current  divides  equally, 
5  amp.  flowing  upward  in  balancer  A,  combining  with  its  down- 
ward flowing  1  amp.  to  give  5—1=4  amp.  in  A,  and  5  amp. 


92  ELECTRICAL  ENGINEERING 

flowing  downward  in  B,  combining  with  its  downward  flowing  1 
amp.  to  give  5+1=6  amp.  in  B.  Current  in  A  flows  similarly 
to  that  in  the  main  generator.  Thus  A  acts  as  a  generator, 
supplying  4  amp.  to  the  load.  Current  in  B  flows  in  the  opposite 
direction;  thus  B  acts  as  a  motor  and  drives  A.  The  difference 
in  current  between  that  in  B  and  that  in  A  is  2  amp.,  which, 
when  multiplied  by  E,  the  voltage  across  B}  gives  2E,  the  power 
required  to  drive  the  balancer  set. 

If  the  generator  voltage  be  assumed  as  200  (that  is,  2^  =  200), 
then  the  generator  output,  or  rating,  if  this  be  full-load,  is 
200  X  36  =  7.2  kw.  The  balancer,  A,  rating,  as  a  generator, 
is  100  X  4  =  0.4  kw. ;  the  balancer  B,  as  a  motor,  receives  input 
=  100  X  6  =  0.6  kw.  The  line  drop  from  the  generator  to  the 
load  is  (40  -+-  30)  r  =  70r,  where  r  is  the  resistance  of  each  of  the 
outer  wires.  The  line  loss,  in  transmission,  is  (402-f-302)r 
=  2500r. 

If  the  entire  load  were  on  the  two-wire  system,  the  current  in 
each  wire  would  be  70  amp.,  the  line  drop,  using  the  same  size 
wires,  would  be  140r,  and  the  line  loss  would  be  (2X  702)r 
=  9800r.  Comparing  the  two-wire  system,  using  the  same  size 
of  outer  wire, 

drop,  three-wire  _   70   _ 
drop,  two-wire  ~"~  140 

loss,  three-wire  _  2500  _ 
loss,  two- wire     ~  9800 

The  middle  wire,  carrying  10  amp.,  has  no  effect  on  the 
total  drop  between  the  outer  wires.  It  does  have  some  effect 
in  slightly  unbalancing  the  voltage  of  the  two  branches  of  the 
system.  Thus,  assuming  the  voltage  across  the  two  machines 
of  the  balancer  to  be  exactly  equal,  which  is  very  nearly  true, 
and  taking  this  voltage  as  E,  the  voltage  across  each  branch  of 
the  load  may  be  found.  Across  the  upper  branch  it  is, 

E  -  40r  -  lOr  =  E  -  50r. 
Across  the  lower  branch  the  voltage  is 

E  -  30r  +  lOr  =  E  -  20r. 

The  amount  of  unbalancing  of  the  voltage  is  therefore 
(E  -  50r)  -  (E  -  20r)  =  30r. 


DIRECT-CURRENT  GENERATORS  93 

To  get  a  concrete  idea  of  the  amount  of  this  unbalancing,  let 

SO 
the   line   drop,    70r,  =  10   per   cent.     Then   30r  =  ^  X  0.1  = 

0.043  =  4.3  per  cent. 

When  the  load  consists  of  lamps  it  is  necessary  that  the 
two  branches  shall  be  sufficiently  well  balanced  to  prevent 
excessive  variation  in  voltage.  This  is  usually  very  easily 
accomplished. 

The  middle  wire  adds,  directly,  a  small  amount  to  the  line 
loss.  In  this  instance,  the  loss  in  this  wire  is  102r  =  lOOr. 

The  total  loss  in  the  system  is  therefore  2600r,  and  the  ratio 

2600 
of  losses  of  the  two  systems  is  QQnn  =  0.265. 


Where  the  percentage  line  drop  or  the  percentage  line  loss  is 
specified,  and  must  be  the  same  with  either  system,  the  ad- 
vantage of  the  three-  wire  system  is  in  the  saving  in  the  cost  of 
copper.  On  that  basis,  let  the  calculations  as  already  carried 
out  for  the  three-wire  system  be  assumed  as  fulfilling  the  re- 
quirements, that  is, 

Line  drop  =  70r. 

Line  loss  (two-wire)  =  2500r. 

The  two-wire  system,  to  give  equal  line  drop  must  be  com- 
posed of  wires  determined  by  the  equation, 

2  X  70  X  r'  =  70r, 

where  r'  =  resistance  of  one  wire  of  the  two-wire  system,  and 
r,  as  before,  is  the  resistance  of  one  of  the  outer  wires  of  the 
three-  wire  system.  Then 


and  each  wire  of  the  two-wire  system  will  be  twice  as  large  as 
each  outer  wire  of  the  three-  wire  system.  Assuming  the  middle 
wire  equal  to  the  outer  wire,  the  two-wire  system  will  require 
four-thirds  as  much  copper  as  the  three-  wire  system. 

Since,  however,  the  variation  in  voltage  .  is  felt  by  all  the 
lamps  on  the  two-wire  system,  while  on  the  other  system  ap- 
proximately one-half  the  variation  in  voltage  is  felt  by  each 
branch,  it  is  more  reasonable  to  calculate  on  the  basis  of  equal 
percentage  drop  in  the  two  systems. 


94  ELECTRICAL  ENGINEERING 

Percentage  drop,  three-wire,      =        =  35  - 


For  equal  percentage  drop,  therefore,  the  two-wire  system 
will  require  eight-thirds  as  much  copper  as  the  three-wire 
system. 

On  the  basis  of  equal  power  loss  in  the  outer  wires, 

9800r'  =  2500r, 


'  r  ~  9800  " 
Adding  the  middle  wire,  equal  to  an  outer  wire,  the  two-wire 

system  will  require  n  055  _u  n  12Y5  =  2.61  times  as  much  copper 
as  the  three-  wire  system. 

Problem  31.  —  What  saving  in  copper  does  the  three-  wire  system  give 
over  the  two-wire  system,  when  the  load  is  balanced,  on  the  basis  of  (a) 
equal  percentage  line  drop,  (&)  equal  line  loss? 

1.  Middle  wire  equal  to  outer  wire. 

2.  Middle  wire  one-half  of  outer  wire. 

3.  Show  that  with  a  balanced  load  no  current  flows  in  the  middle  wire. 
Problem  32.  —  One  hundred  60-watt  tungsten  lamps  are  to  be  supplied 

with  power  at  3  per  cent,  line  loss.  The  line  length  is  600  ft.  Lamp 
voltage  is  120.  The  neutral  wire  is  to  be  one-half  the  cross-section  of 
each  outer  wire. 

Find  the  size  of  the  required  wires,  and  show  that  the  weight  of  copper 
is  approximately  190  Ib.  Show  that  on  the  two-wire  system  610  Ib.  would 
be  required. 

Feeder 


/     T        f                                   Trolley  Wire 

4 

rf 

/ 

E5 

iiUj 

Rail                                         EEL 

COJQ 

FIG.  73. 


Boosters. — Generators  are  frequently  connected  in  series  for 
the  purpose  of  regulating  the  voltage  and  equalizing  it  along  a 
line  in  which  there  is  considerable  voltage  drop. 

Fig.  73  shows  a  simple  arrangement  of  a  street  railway  circuit 
in  which  a  booster  is  used.  The  generator,  G,  supplies  power 


DIRECT-CURRENT  GENERATORS 


95 


to  the  system,  including  that  delivered  directly  to  the  car  and 
that  used  in  driving  the  motor  M .  The  motor  and  booster  form, 
usually,  a  directly  connected  set.  One  terminal  of  the  booster  is 
connected  to  the  trolley  wire  at  the  station,  the  other  is  con- 
nected through  a  heavy  feeder  to  some  distant  point  on  the 
trolley  wire. 

As  an  example  of  the  effect  of  using  a  booster,  consider  the 
following : 

Problem  33. — A  trolley  line  3  miles  long  is  supplied  with  power  by  a 
generator  at  600  volts.  The  trolley  wire  is  of  No.  00  B.  &  S.  wire,  having  a 
resistance  of  0.4  ohm  per  mile.  The  rail  return  has  a  resistance  of  0.05 
ohm  per  mile.  A  feeder,  consisting  of  three  No.  0000  B.  &  S.  wires,  of 
0.087  ohm  per  mile,  extends  from  the  station  to  a  point  2  miles  distant, 
where  it  connects  with  the  trolley  wire.  The  booster  voltage  is  maintained 
at  40.  Find  the  voltage  on  the  car  as  it  proceeds  from  the  distant  end  of 
the  line  toward  the  station,  assuming  that  the  current  taken  is  at  all  times 
200  amp. 

Solution. — It  will  be  of  interest,  first,  to  determine  the  voltage  on  the  car 
at  the  distant  end  when  the  booster  is 
disconnected. 

Drop  in  the  trolley  wire  is  200  X  0.4 
X  3  =  240  volts. 

Drop  in  rail  =  200  X  0.05  X  3  =  30    «»v. 
volts. 

.'.  Voltage  on  car  without  booster  = 
600  -  270  =  330. 

This  is  to  illustrate  the  necessity  of  doing  something  to  improve  the  regu- 
lation of  the  line.  With  the  booster  connected,  the  problem  becomes  one 
for  the  application  of  KIRCHOFF'S  laws. 

The  circuit  is  represented  diagrammatically,  in  Fig.  74,  for  the  case  of  the 
car  at  the  end  of  the  line.  Arrows  indicate  arbitrary  directions  of  flow  of 
current.  Let  the  voltage  on  the  car  be  denoted  by  E. 

By  KIRCHOFF'S  laws, 

ii  +  12  =  200 
and 

40  -  ;2r2  +  iiri  =  0, 
whence,  eliminating  i\  between  the  equations, 

200r1  +  40 

iz  = : — 


74 


Substituting  values, 


ri  =  0.4  X  2  =  0.8  ohm 
r2  =  0.087  X  2  =  0.174 

r3 
r0 


=  0.4 

=  0.05  X  3  =  0.15 


200  X  0.8  +  40 
-^8Tol74-=2 
200  -  205  =  -  5  amp. 


96  ELECTRICAL  ENGINEERING 

The  equation  of  the  mesh  composed  of  the  generator,  ri,  r3,  E  and  r0  is 

600  =  iiri  +  200r3  +  E  +  200r0. 
Substituting  values  and  solving  for  E, 

E  =  600  +  4  -  80  -  30  =  494  volts. 

Thus,  there  is  a  total  drop  of  106  volts  instead  of  270  volts  without  the 
booster. 

Now  let  the  car  be  at  the  point,  O,  where  the  feeder  joins  the  trolley  wire. 
Evidently  the  same  equations  hold,  and  z'2  =  205  amp.,  ?i  =  —  5  amp. 

r0  is  now  0.05  X  2  =0.1  ohm. 
The  mesh  equation  is  now 

600  =  iiri  +  E  +  200r0. 
/.  E  =  600  +  4  -  20  =  584  volts. 

When  the  car  is  at  a  point  1  mile  from  the  generator,  the  current  and 
voltage  equations  are : 

^  +  12=  200 
and 

40  -  i,  (r,  +  0  +  ti  £  =  0. 
Solving  for  i2  gives 

i»  =  123  amp. 
and 

ii  =  200  -  123  =  77  amp. 

The  mesh  equation  of  voltage  is 

600  =  ii  ^  +  E  +  200r0, 

where  r0  is  now  0.05  ohm.  « 

/.  E  =  600  -  31  -  10  =  559  volts. 

Thus,  it  is  seen  that,  by  this  simple  connection  of  the  booster  to  a  point 
chosen  more  or  less  at  random,  the  voltage  has  been  rendered  much  more 
nearly  uniform  than  it  would  be  without  the  booster. 

Problem  34. — As  a  further  study  of  the  booster  problem,  consider  that  in 
the  above  case  the  feeder  is  to  be  connected  to  the  trolley  wire  at  two  points, 
namely,  at  1.5  and  2.5  miles  from  the  generator. 

Find  the  voltage  on  the  car  at  each  half-mile  point,  and  plot  against 
distance  from  the  generator. 


CHAPTER  XIII 
DIRECT-CURRENT  MOTORS 

If  two  shunt  generators  connected  in  parallel  supply  power  to  a 
certain  load,  as  in  Fig.  75,  the  division  of  the  load  between  the 
generators  will  depend  upon  their  respective  degrees  of  excitation. 

By  weakening  the  field  of  No.  1,  it  will  take  less  of  the  load 
until,  by  continued  weakening,  it  takes  none  at  all  and  finally 
receives  current  from  No.  2,  thus  being  run  as  a  motor. 

With  a  change  in  direction  of  flow  of  current  in  the  armature 
comes  a  change  in  the  direction  in  which  the  armature  tends  to 
rotate  due  to  its  current,  the  direc- 
tion of  the  field  remaining  constant 
in  shunt  machines. 

As  a  generator  the  rotational  force 
of  the   armature  is   counter  to    the 

actual  direction  of  rotation  which  is 
,  .,      ,  .   .  .          TT  FIG.  75. 

due  to  tne  driving  engine.     However, 

the  actual  direction  of  rotation  does  not  change  when  the 
machine  ceases  to  act  as  a  generator  and  becomes  a  motor. 

With  the  series  generator,  reversal  of  the  armature  current 
also  reverses  the  field.  To  obtain  a  generator  action  from  a  series 
motor,  therefore,  requires  reversal  of  rotation. 

It  has  been  shown  that,  with  generators,  a  forward  shift  of 
the  brushes  increases  the  armature  demagnetization. 

With  a  shunt  motor  the  armature  currents  are  reversed,  the 
armature  ampere-turns  are  reversed,  and  the  effect  of  the  arma- 
ture, in  shifting  the  resultant  flux,  is  consequently  reversed. 
Therefore,  the  brushes  of  a  motor  require  to  be  given  a  backward 
shift.  The  effect  of  a  backward  shift  on  a  motor,  like  the  for- 
ward shift  on  a  generator,  is  to  increase  the  armature  demag- 
netizing ampere-turns. 

With  direct-current  motors,  the  impressed  e.m.f.  is  the  sum  of 
the  counter  e.m.f.  and  the  ir  drop. 

Thus,  the  fundamental  equation  is 

E  =  Ei  +  ir 
7  97 


98  ELECTRICAL  ENGINEERING 

where  E  =  impressed  e.m.f., 
Ei  =  counter  e.m.f., 

i  =  current, 
and 

r  =  resistance  of  armature,  brushes,  etc. 

The  generator  equation  (20)  also  applies  to  the  counter  e.m.f., 
since  the  counter  e.m.f.  is  the  generated  e.m.f.  of  the  motor  due 
to  the  rotation  of  its  armature  conductors  in  the  field. 


where 

,         4< 
=  105' 

Substituting  this  value  of  Ei} 

E  =  kf<f>  +  ir  (24) 

whence 


f  =  frequency.     To  transform  frequency  to  speed, 

r.p.m.       p 
J  '        60       *\i* 

where 

p  =  number  of  poles. 

For  ordinary  operation, 
pi 
f  =  7—'  approximately. 

There  are  three  ways  of  changing  the  speed  of  a  direct-current 
motor:  (1)  by  changing  E,  the  impressed  voltage;  (2)  by  changing 
<f>  by  means  of  a  field  rheostat  ;  (3)  by  changing  </>  by  shifting  the 
brushes. 

Shifting  the  brushes  is  not  an  effective  means  of  speed  regula- 
tion since  it  introduces  trouble  from  sparking  at  the  brushes. 

Types  of  Direct-current  Motors.  —  The  principal  types  of 
direct-current  motors  are  known  as  shunt,  series,  cumulative- 
compound,  in  which  the  series  and  shunt  turns  act  in  the  same 
direction,  and  differential-compound,  in  which  the  two  field 
m.m.fs.  are  arranged  to  oppose  each  other. 


DIRECT-CURRENT  MOTORS  99 

Speed  Characteristics  of  Direct-current  Motors.  —  These  are 
curves  between  speed  and  load,  the  latter  being  the  independent 
variable.  To  determine  the  effect  of  load  upon  speed,  in  the  case 
of  shunt  motors,  it  is  seen  from  Eq.  (25), 

_  E  -  ir 

*  =  ~ 


that  an  increased  ir  drop  tends  to  reduce  the  speed.  It  has  also 
been  shown  that  <f>  is  reduced  by  armature  reaction,  in  pro- 
portion, roughly,  to  the  load.  Therefore,  for  shunt  motors,  the 
relation  between  the  armature  reaction  and  the  ir  drop  will  de- 
termine whether  the  motor  will  speed  up  or  slow  down  with  an 
increase  of  load.  In  general,  if  the  magnetization  of  the  field 
extends  above  the  knee  of  the  saturation  curve,  the  motor  will 
slow  down,  while  below  the  knee  the  motor  will  speed  up.  Evi- 
dently, a  degree  of  magnetization  might  be  obtained  which  would 
result  in  practically  constant  speed. 

The  cumulative-compound  motor  slows  down  with  increase 
of  load,  since  the  effect  of  the  series  turns  is  to  strengthen  the 
field. 

The  differential  motor  speeds  up 
with  increasing  load,  due  to  the  op- 
position of  the  series  and  shunt  field 
m.m.fs. 

The  series  motor  speed  is  governed 
almost  entirely  by  its  field,  which  is 
nearly  proportional  to  the  load  cur- 
rent. At  light  loads,  the  speed  be- 
comes high  and  the  operation  of  the  motor  is  unstable.  In  Fig. 
76  is  shown  a  set  of  speed  characteristic  curves. 

The  student  should  be  able  to  establish  the  general  speed 
equations  and  derive  curves  for  each  type  of  motor. 

Power  and  Torque.  —  Power  input  to  the  motor  is  obtained  by 
multiplying  Eq.  (24)  by  i,  thus 

Wi  =  Ei  =  Ej  +  i*r  =  kf<(>i  +  i2r. 

In  this,  equation  E#  represents  the  output  of  the  motor  in 
mechanical  work,  including  bearing  friction  and  windage;  i2r  is 
the  power  lost  as  heat  developed  in  the  armature. 

Expressed  in  horsepower,  the  output  is 


Load 

FIG.  76. 


100 


ELECTRICAL  ENGINEERING 


Horsepower  may  also  be  expressed  as 

2irRnF 
P'  ~  33,000 

where  R  =  radius  of  armature  in  feet, 

n  =  revolutions  per  minute, 

F  =  force  in  pounds  on  the  armature  conductors. 
2wn  =  co  =  angular  velocity, 
and   RF  =  T  =  torque. 

Thus 
and 


746      33,000 

33,000  Ed 
27rnX746' 

But  output  is  also,  by  (24),  kf<f>i. 

.        ^  33,000  kfoi 
=  2irn  X  746 ' 

Also,  since  /  = 
T  = 


where  p  =  number  of  poles, 
0.0587  kp4>i  = 


120 
33,000  kp<t>i 


27rX  746X120 
This  expression  may  be  reduced  still  further,  since 


Thus,  for 


where  t  =  number  of  turns  in  series  on  the  armature. 

a  motor  of  p  poles  and  t  turns  in  series, 

T  =  0.2348  tp<t>i  X  10~8  ft.-lb. 
Torque  Characteristics.  —  From  the  above  equation  of  torque 

it  is  possible  to  construct  curves  showing  torque  variation  with 

load  current.  It  is  necessary,  however, 
to  be  able  to  find  the  value  of  0  in 
each  case.  With  shunt  motors  </>  is 
nearly  constant,  and  torque  is  therefore 
nearly  proportional  to  current.  With 
series  motors  <f>  increases  with  i,  and 
torque  therefore  goes  up  as  the  square 
of  the  current,  approximately.  l  Fig.  77 
gives  a  set  of  torque  characteristics  for 

the  four  types  of  direct-current  motor. 

1  When  the  field  core  becomes  saturated,  increase  of  current  does  not 
produce  much  increase  of  flux.  Under  heavy  loads,  therefore,  the  torque  of 
a  series  motor  increases  more  nearly  in  direct  proportion  to  the  current. 


Current 

FIG.  77. 


DIRECT-CURRENT  MOTORS  101 

Problem  35.  —  Direct-current  motors  and  generators  being  entirely 
similar  as  respects  fundamental  equations,  armature  reaction,  etc.,  it  is 
thought  best  to  submit  to  the  student  the  problem  of  the  direct-current 
shunt  motor  instead  of  presenting  it  here  in  detail.  Let  the  generator 
whose  design  was  worked  out  in  Chap.  X  be  now  considered  as  a  shunt 
motor.  The  series  turns  will  then  be  disconnected.  With  250  volts  im- 
pressed on  the  armature  and  maintaining  constant  shunt  field  amp.  -turns 
of  7500,  let  it  be  required  to  calculate  the  speed  and  plot  its  values  against 
those  of  the  load  current. 

Choose  current  values  of  0,  1000,  2000,  3000  amp.  Assume  a  constant 
brush  shift  of  15°. 

The  fundamental  speed  characteristic,  Eq.  (20),  has  been  found  to  be 

E  -  ir 

'--*r- 

number  of  poles  X  r.p.m.        pn 
where  2X60  =  120' 

E  =  impressed  voltage, 
r  includes  both  armature  and  brush  resistance. 


where  t  =  number  of  turns  per  pole  on  the  armature  and  $  is  the  flux  cutting 
the  armature  conductors.     For  this  last  it  is  sufficiently  exact  to  assume 

$  at  load  amp.  -turns  at  load 

0  at  no-load  ~~  amp.  -turns  at  no-load 
(See  armature  reaction,  Chap.  X.) 

Problem  36.  —  The  same  problem  as  the  preceding  should  now  be  worked 
out,  using  (1)  E  =  270  volts,  (2)  E  =  220  volts. 

Question.  —  What,  in  general,  is  the  effect  on  shunt  motors  of  increasing 
or  lowering  the  terminal  voltage,  as  regards  (a)  speed,  (6)  torque,  (c)  output, 
(d)  efficiency? 

Problem  37.  —  Let  the  above  motor  be  calculated  as  a  differential-com- 
pound machine,  the  series  ampere-turns  to  be  so  adjusted  as  to  give  the 
same  field  strength  at  full-load  as  at  no-load. 

Plot  speed  vs.  armature  current  for  impressed  voltage  E  =  250. 

Problem  38.  —  Same  as  37  only  the  motor  is  to  be  connected  as  cumulative 
compound. 

Problem  39.  —  If,  now,  the  entire  field  strength  were  determined  by  the 
series  turns,  so  that  at  full-load  there  should  be  10,427  series  amp.-turns,1 
calculate  and  plot  the  speed  for  variation  of  load. 

Series  field  circuit  resistance  may  be  taken  as  0.00134  ohm. 

Problem  40.  —  In  problems  35,  37,  38  let  the  speed  be  maintained  con- 
stant by  variation  of  the  shunt  field  current.  Let  this  speed  be  that  of  the 
shunt  motor  at  no-load  (E  =  250). 

Plot  curves  between  field  current  and  load  current. 

Problem  41.  —  Show  how  to  obtain  constant  speed  by  shifting  the  brushes, 
and  work  out  numerically,  as  far  as  possible,  the  case  of  the  shunt  motor. 
Plot  a  curve  between  degrees  of  brush  shift  and  load  current. 

1  Same  as  required  for  the  generator  at  full-load,  Chap.  X. 


CHAPTER  XIV 


N 


FIG.  78. 


THEORY  OF  THE  BALLISTIC  GALVANOMETER 

This  particular  type   of  galvanometer  is  of  importance    in 
magnetic  measurements,  especially  in  the  determination  of  the 

hysteresis  loop. 

It  consists,  usually,  of  a  coil  of 
fine  wire  wound  upon  a  steel  cylin- 
*          der,  freely  suspended  between  the 
poles  of  a  magnet  as  illustrated  in 

Fig.  78. 

It  has  been  shown  that  the  force 
exerted  on  a  wire  carrying  current, 
when  placed  in  a  field  perpendicular  to  the  lines  of  flux,  is 

F  =  Eli  dynes, 

where  i  is  current  in  abamperes,  I  is  length  of  wire  in  centimeters 
and  B  is  flux  density  in  lines  per  square  centimeter. 

If  the  wire  is  one  side  of  a  rectangular  loop,  then  the  turning 
couple  of  the  loop  is 

C  =  2pBli  dyne-cm. 

When  the  loop  is  displaced  by  an  angle, 
0°,  from  the  direction  of  the  flux  lines  (Fig. 
79),  the  couple  is 

C0  =  2pBli  cos  6  =  ABi  cos  6, 
where  A  =  2pl  =  area  of  the  loop. 

When  the  current  is  sent  through  the 
loop,  the  action  of  the  couple  produced  is 

to  turn  the  loop  through  an  angle,  0.  In  order  to  oppose  this 
action,  a  spring  is  so  attached  to  the  loop  as  to  introduce  an  op- 
posing couple,  k&,  which  balances  the  swing  of  the  loop.  Then 

ke  =  ABi  cos  0 
and 

K0 
AB  cos  6 

where  k  is  a  constant  of  the  spring. 

102 


Force 


FIG.  79. 


THEORY  OF  THE  BALLISTIC  GALVANOMETER    103 

For  small  angles,  6  =  sin  6.     Substituting  this, 
k  sin  6  k 


cos  0 


tan 


Thus,  the  current  in  the  loop  is  directly  proportional  to  the 
tangent  of  the  angle  of  deflection;  hence,  the  "  tangent" 
galvanometer. 

For  small  angles,  also,  6  =  tan  6. 


Thus,  the  galvanometer  •  may  be  used  as  an  ammeter  to 
measure  directly  the  current,  so  long  as  the  angle  of  deflection  is 
kept  small. 

In  the  ballistic  galvanometer  the  moving  part  is  designed 
to  have  much  inertia,  so  that  its  natural  period  of  vibration 
shall  be  long  in  comparison  with  the  time  of  change  of  the  flux 
to  be  measured.  Thus,  a  change  of  flux,  produced  in  a  sample 
of  iron  under  test  by  altering  the  number  of  ampere-turns  on 
the  iron,  will  take  place  before  the  loop  can  move,  that  is,  while 
0  =  0  and  cos  0=1. 

The  couple  on  the  loop  is  then 

C  =  ABi, 

which  causes  the  loop  to  accelerate. 

Therefore,  ABi  is  the   couple   of  angular  acceleration,  and 


where  /o  =  moment  of  inertia  of  the  moving  element,  and  o>  = 
angular  velocity. 

But  idt  is  the  quantity  of  electricity  flowing  in  any  time,  dt. 
Therefore  the  total  quantity 


§   ...         /o    |    ,          h  <*> 

J  ldt  =  AB J  d"  =  ~KB 


(26) 

AD 

where  a>0  is  the  final  velocity  attained. 

The  deflection  is,  however,  limited  by  k0,  the  torsion  of  the 
spring.     The   work   done   in   overcoming  this  torsion  is   then 


W  =      kOdd  = 
where  00  is  the  maximum  deflection. 


=  (k 


104  ELECTRICAL  ENGINEERING 

Solving  this  equation, 


I/O 

t/o  —  ^OA/ 
But  by  (26), 

0)fl 

Substituting  this  value  of  co0, 


whence, 

Q  =     ^^  (27) 


In  this  equation  all  terms  are  constant  except  00,  the  maximum 
deflection  of  the  loop. 

If,  now,  the  change  of  flux  is  d<f>  in  a  time  dt,  the  e.m.f  .  induced 
in  the  coil  surrounding  this  flux  is 

N_  d$  _    . 
108  dt  ''    IT) 

where  N  is  the  number  of  turns  of  the  coil,  r  is  the  resistance 
of  the  circuit,  and  i  is  the  current  set  up  in  the  circuit. 
Then,  transposing, 

Nd<t> 


idt  =  - 


r  X  108' 


The  total  quantity  of  electricity  set  flowing  by  the  change 
of  flux  is  then 


r  X  108  L  r  X  108  ' 

whence,  from  (27), 

Qr  X  108  _  60  r  X  108\/J7fc 
N  ABN 

is  the  maximum  value  of  the  flux. 

There  is  thus  a  direct  relation  between  flux  and  maximum 
deflection,  and  0o  is  therefore  a  measure  of  the  flux. 


CHAPTER  XV 

VECTOR  REPRESENTATION  OF  ALTERNATING- 
CURRENT  WAVES 

In  Chap.  VIII  the  graphical  relationships  of  the  waves  of 
voltage  and  current  in  an  alternating-current  inductive  circuit 
have  been  developed,  and  the  values  and  meaning  of  average 
and  effective  values  of  a  sine  wave  have  been  discussed. 

The  waves  of  Fig.  37  may  also  be  represented  as  vector  pro- 
jections of  their  maximum  values  on  the  vertical  axis,  as  shown 
in  Fig.  80.  Since  i  =  Im  sin  0  the  length  of  ^  r 

the  current  vector  is  taken  as  Im  and  the 
value,  i,  at  any  instant,  is  the  vertical  pro- 
jection of  Im  as  it  uniformly  rotates,  at  speed 
2-7T/  about  the  origin.  The  vectors  all  have  FIQ  ^ 

the  same  speed  of  rotation  so  that  their  re- 
lations to  each  other  are  constant.  Hence  their  position  in 
space  at  any  desired  instant  may  be  chosen.  Let  that  instant 
be  when  0  =  o,  in  Fig.  37.  Then  i  =  Im  sin  0  =  o,  and  Im 
must  be  laid  off  horizontally.  rlm,  the  maximum  value  of  the 
e.m.f.  consumed  by  the  resistance,  since  it  is  in  time-phase  with 
Im,  is  also  laid  off  horizontally;  xlm,  the  maximum  value  of  the 
e.m.f.  consumed  by  the  inductive  reactance,  x,  is  90°  ahead  of 
Im,  and  is  therefore  laid  off  vertically  upward.  Thus  xlm 
is  positive  maximum  when  Im  is  at  zero,  becoming  positive. 
rlm  and  xlm  may  now  be  added  vectorially,  giving  ZIm  or 
Em  which  is  the  maximum  value  of  e.  Em  is  seen  to  be  placed 

at  an  angle  a  ahead  of  Im,  such  that  tangent  a  =  -y—  =  —  • 

•  J-  m  ' 

This  relation  is  also  of  fundamental  importance.     The  numerical 
value  of  Em  is  obtained  by  the  relation 

Em  =  Vlm2r*  +  ImV  =  Im  Vr2  +  z2 
The  quantity  \/r*  +  x2  is  called  the  impedance  and  is  denoted 

by  the  letter  z. 

105 


106  ELECTRICAL  ENGINEERING 

Problem  42. — Draw  the  vectors  of  e.m.f.  and  current  of  problem  28, 
Chap.  VIII,  and  show  that  the  angle  of  lag  of  current  behind  e.m.f.  is 
38°  40'. 

In  the  representation  of  waves  by  vectors,  the  vectors  are  not, 
in  reality,  moved,  but  their  relative  positions  in  space  are  con- 
sidered. Since  no  rotation  is  required,  they  may  therefore  be 
drawn  in  length  equal  to  their  effective  values,  and  this  is  the 
common  method  of  representation. 

Also,  since  (7mz)2  =  (Imr)*  +  Umx)2  (28) 

£2    =    r2    _|_  3.8  (29) 

and  the  vector  relationship  holds  for  (29)  as  for  (28).  There 
can  be  constructed  what  is  known  as  the  im- 
pedance triangle,  Fig.  81,  in  which  a  =  r,  b  = 

b       x 
x,  c  =  z,  and  tan  a  =  -  =  — 


Thus, 


FIG'  8L 

x  =  z  sin  a. 

Substituting  these  values  in  (19),  e  =  Im(r  sin  6  +  x  cos  6), 
gives, 

e  =  7m2(sin  B  cos  a  +  cos  6  sin  a) 

=  Imz  sin  (8  +  a), 

or,  substituting  for  6,  its  equivalent,  coZ, 

e  =  Imz  sin  (ut  +  a),  (30) 

in  which  at  is  a  variable  angle  depending  on  t,  and  a  is  a  constant 
angle  determined  by  the  relative  values  of  x  and  r.  Eq.  (30) 
shows  that  e,  like  i,  is  a  sine  wave  quantity,  but  that  there  is  a 
constant  angular  or  phase  difference,  a,  between  them,  a  is 
called  the  angle  of  lead  or  lag,  depending  on  whether  it  is  posi- 
tive or  negative. 

The  relations  indicated  in  Fig.  81  may  also  be  expressed  by  the 
notation  of  complex  quantities. 
Thus, 

c  =  a  -f  jb. 

The  addition  of  the  letter  j  to  the  equation  simply  means  that 
the  quantity,  6,  is  to  be  drawn  vertically  upward.  If  it  were 
—  j,  b  would  be  drawn  vertically  downward.  A  dot  is  put  under 


ALTERNATING-CURRENT  WAVES  107 

c  which  means  that  c  is  dealt  with  as  a  vector  quantity.    Without 
the  dot,  the  scalar  or  numerical  value  of  c,  only,  is  meant. 
Thus,  _ 

c  =  \/a2  +  V. 

Problem  43.  —  Show  graphically  that  3-J3  is  a  vector  of  length  4.24, 
which  makes  an  angle  of  —  45°  with  the  horizontal  axis.  Show  that  a  vector 
of  length  12,  at  angle  120°,  is  represented  by  the  expression,  —  6  +  j  10.4. 

Calculate  and  draw  the  following  vectors  :  c  =  3  —  j'2,  c  =  4  +  j,  c  = 

-2  +  j3,  c  =  -4  -J2. 


j  also  means  a  rotation  of  90°  in  the  positive  or  counterclock- 
wise direction.  If  the  vector,  a,  is  multiplied  successively  by  j, 
several  times,  its  direction  is  shown  as  follows: 


Vector                                                       Angle 
....                    0° 

90° 

Ju  

jja  —  —  a, 

180° 

270° 

iiiia  =  a.  . 

.  .  360°  =  0° 

Thus  may  be  written, 
whence 


or,  j  is  identical  with  i,  used  commonly  in  mathematics  to  denote 
imaginary  quantities.1 

If  it  is  desired  to  rotate  a  through  30°,  we  can  write 

a  =  a  cos  30°  -f  j  a  sin  30°. 


To  rotate  «°  correspondingly,  ^~~°      ' a  8in30° 

a  =  a  cos  a°  +  j  a  sin  a°, 

/  o    i  o\  FIG.  82. 

=  a  (cos  a    -f  j  sin  or). 

Suppose  a  is  first  rotated  30°,  then  60°  more.     Then  a  =  a  (cos  30° 
+  j  sin  30°)  (cos  60°  -f-  j  sin  60°). 

Problem  44. — Prove  that  this  double  rotation  results  in 

a  =  ja 

Consider  the  simple  case  of  alternating  current  in  an  inductive 
resistance,  Fig.  83,  where  current,  /,  resistance,  r,  and  reactance, 

1  In  electrical  engineering  j  is  used  instead  of  i,  because  i  is  used  to  denote 
current. 


108 


ELECTRICAL  ENGINEERING 


x,  are  known.  /  is  chosen  as  the  zero  vector.  Then  I  =  i. 
Frequently  it  is  well  to  choose  as  the  zero  vector,  or  vector  drawn 
at  0°,  some  known  quantity.  In  order  to  determine  the  positions 
of  the  vectors  of  electromotive  force,  etc.,  with  respect  to  the  zero 
vector,  there  are  two  rules,  previously  brought  out,  which  are 
important  to  remember: 

Rule  I. — The  e.m.f.  consumed  by  resistance  is  in  time-phase 
with  the  current,  and  in  the  same  direction. 

Rule  II. — The  e.m.f.  consumed  by  inductive  reactance  is  in 
time-phase  90°  ahead  of  the  current 

By  these  rules  may  be  drawn  the  vector  diagram,  Fig.  84,  in 
which  the  vector  sum  of  ix  and  ir  is  iz,  which  is  the  total  electro- 
motive force  consumed. 


FIG.  83. 


FIG.  84. 


This  electromotive  force  consumed,  or  vector  E,  numerically 
equal  to  iz,  is  represented  by  the  relation 

E  =  ir  +  jix  —  i(r  +  jz). 


The  impedance  is  thus  expressed  as  r  +  jx,  and  it  is  a  vector  of 
magnitude  z  =  \/r2  +  x2,  and  the  angle  between  the  impedance 
and  the  resistance  is  defined  by  the  relations. 


and 


Z  COS  a 


tan  a  =  -• 
r 


The  e.m.f.  consumed  in  the  circuit  is,  in  general, 
E  =  IZ  =  (i±ji'}  (r+jx). 

The  current  may  or  may  not  be  chosen  as  the  zero  vector.  If  it  is 
so  chosen,  /  =  i.  If  not,  then  /  =  i  ±  ji'}  where  i'  is  the  wattless 
component  of  the  current. 

The  impedance  is  always  z  =  r  +  jx. 

Assigning  positive  or  negative  values  to  the  wattless  component 
if,  we  may  write,  in  any  case, 

I  =  i+  ji'. 


ALTERNATING-CURRENT  WAVES 


109 


It  should  be  remembered  that  a  leading  component  requires  a 
+  sign,  and  a  lagging  component  requires  a  —  sign. 
Therefore,  E  =  IZ  =  (i  +  ji')  (r  +  jx) 

=  ir  —  i'x  +  j(i'r  +  ix) 

If  the  current  is  taken  as  the  zero  vector,  then 
E  =  i(r  +  jx) 

In  the  general  expression  (31),  an  arbitrary  zero  line  is  chosen, 
as  in  Fig.  85.  In  the  simpler  case  (32),  the  direction  of  /  is 
chosen  as  the  zero  line,  whence  I  —  i  and  i'  =  0,  and  the  vector 
diagram  becomes  that  of  Fig.  86. 


(31) 


(32) 


ix 


FIG.  85. 


%r 

FIG.  86. 


Problem  46.  —  One  ampere  flows  in  a  circuit  of  1  ohm  resistance  and  a  vari- 
able reactance.  Plot  curves  of  Ir,  Ix,  Iz  drops  and  phase  angle  against  x, 
when  x  varies  from  0  to  5  ohms.  Take  /  as  the  zero  vector.  Then  7  = 
i  =  1. 


Solution.  — 


Tabulating : 


•'  tan  a 


X 

0 

0.5 

1 

2 

3 

4 

5 

ix  

0 

0.5 

1 

z  

1 

1.12 

iz 

1 

1  12 

x 

0 

0  5 

r 

- 

a°  

0 

26°  35' 

The  blank  spaces  may  be  filled  in  by  the  student. 

Consider  the  same  case,  Fig.  79,  but  with  E  known  and  / 
unknown.  E,  then,  may  conveniently  be  chosen  as  the  zero 
vector,  and 

I  =  e_  =       e  e(r-jx) 

z       r  +  jx      ( 


(33) 


110 


ELECTRICAL  ENGINEERING 


The  last  expression  of  (33)  is  obtained  in  accordance  with  a  third 
rule,  as  follows: 

Rule  HI.  —  Never  allow  an  equation  to  remain  with  a  complex 
denominator.     Thus  (33)  becomes 


where 


g 


(34) 


(35) 


FIG.  87. 


g  +  jb  —  F,  is  called  the  admittance-;  g  is  called  the  conductance, 
and  b  the  susceptance  of  the  circuit.  The  diagram  of  currents 
may  now  be  drawn  to  correspond  with 
Fig.  87,  for  e.m.fs.,  in  which  eg  is  the  com- 
ponent of  the  current  in  phase  with  e,  that 
is,  it  represents  energy  expended,  and  —  eb 
is  the  component  90°  behind  e,  called  the 
reactive  or  wattless  component  because 
it  does  not  represent  any  expenditure  of 
energy. 
The  power  input  to  the  circuit  is  then 

Power  input  =  e  X  eg  =  ezg, 
and  this  is  found  equal  to  I2r. 

The  numerical  value  of  the  current  =  I  =  e\/g2  +  62. 

Problem  46. — Let  E  =  e  =  1;  x  =  1;  r  varies  from  0  to  10.  Plot  curves 
of  /  vs.  r,  and  I2r  vs.  r. 

Calculate  the  maximum  value  of  the  power  loss  and  find  the  value  of  the 
resistance  which  gives  the  greatest  dissipation  of  power. 

Plot  the  3  current  waves,  that  is,  the  power  current,  eg.,  wattless  cur- 
rent, eb,  and  total  current,  ey,  for  the  condition  of  maximum  power  loss. 

Solution. — 

e  1  1 


Vr2  -f  x2       \/l  +  r2       *  - 

r                x             1 

v  =  -z',°  =  -£-,;  y  =  -• 

Tabulating  : 

r 

0 

2 

4 

6 

8 

10 

Z  

1 

2.27 

4.12 

7  

1 

0.44 

0.242 

/«  

1 

0  194 

7«r..  '..... 

0 

0.388 

* 

ALTERNATING-CURRENT  WAVES  111 


Power,  W  =  Pr  =  -z  r 

*?-.. 


For  maximum  power  -p-  =  0.     .'.  x2  —  r2  =  0,    and    r2  =  x2.     .'.    r2  =  1, 
r  =  1,  and  W  =  1  X  j-qjj  =  0.5  watt. 

To  get  current  waves  for  maximum  power  loss,  then 

r  =  1;  x  =  1;  Z  =  1.41. 

e    _  _e(r  - j x)  _er        ,  ex  _ 


where  ^  =  -^-2and  6  =  —  ^* 

The  effective  values  of  current  are,  therefore, 

eg  =  1  X  K  =  0.5,  in  phase  with  e, 
jeb  =-je~=-jlX^  =  -jO.5,    or    0.5,    90°  behind  e, 

e(g  +jb)=eY=^  =  —^  -  0.707,  lagging  behind  e,  by  an  angle  tan"1 
&       V2 


Maximum  values  are  ^TO  =  \/2^  =  1.41, 

(Eg)m  =  0.707;  (J^6)w  =  0.707;  (EY)m  =  1. 


Circuit  of  Resistance  in  Series  with  an  In- 


ductive   Impedance.  —  The   impressed  e.m.f., 

E,  of  the  circuit  is  known,  also  the  resistance,     E 

r,  and  impedance,  Zi  =  rl  +  jxi  (Fig.  88).     #      I 

is  taken  as  the  zero  vector.     Then,  FlG   88 


where 

r0     ,  Xi      „ 

Q  —  v~2>  o  =  --  ^-2;  Z0 

and  r0  =  r  -j-  ri. 

The  drop  across  the  impedance,  Zi,  is 
#1  =  7  Zi  =  e(g  +  j6)  (7*1  +  ji 

«(s^i  +  n. 


where  a0  =  0n  —  60; ij  60  =  gxi  + 


112  ELECTRICAL  ENGINEERING 

Problem  47. — In  the  above  circuit,  Fig.  88,  let  E  =  10,  r  =  1,  n  =  0.5, 
Xl  =  2. 

Draw  the  vector  diagram  and  waves  of  e,  EI  and  /. 

Circuit  of  Two  Inductive  Impedances  in  Parallel. — Let  E,  Z\  and  Z2  be 
known  (Fig.  89).  To  determine  7,  /i  and  /2. 


FIG.  89. 

We  have: 

/i  =  eYl}  J2  = 

I    -li  +  J,  -a(Fi-+  7,).    ' 

Or, 

/2  =  e(flfi  +j'6i)i 


62)  =  e(G 
where 


G  =  0i  +  gt,  B  =  61  +  62. 

D 

Tan  a  —  -^->  gives  the  phase  relation  of  /  and  e. 

Problem  48.—  In  the  circuit  of  Fig.   89,  let  E  =  1,   n  =  1,   xl  =  0.5, 
r2  =  2,  x2  =  2. 
Draw  vector  diagrams  and  waves  of  E,  Eg,  Eb,  and  /. 


CHAPTER  XVI 

THE    SYMBOLIC    METHOD    IN    TRANSMISSION    LINE 
CALCULATION 

KENNELLY  AND  STEINMETZ  have  introduced  the  so-called 
symbolic  method  of  representing  electrical  relations. 

This  method  is  neither  vector  analysis  nor  quaternions,  but 
is  in  many  ways  similar  to  both.  It  enables  the  use  of  simple 
algebraic  transformation  when  dealing  with  vector  quantities  of 
the  same  rate  of  rotation  or  frequency.  Thus,  it  is  directly 
applicable  when,  for  instance,  multiplying  a  current  by  an 
impedance,  since  the  resultant  e.m.f.  is  of  the  same  frequency  as 
the  current.  But  when  multiplying  current  and  e.m.f.,  it  is 
applicable  only  after  some  modification,  since  the  product 
represents  power,  which  is  a  vector  of  double  frequency. 

Addition. — Let  there  be  two  vectors, 

a\  +  J&2,  and  bi  -\-  jb2,  and  let  their  sum  be  a  vector  C. 
Then, 

C  =  ai  +  ja2  +  61  -f  jb2  =  ai  +  bi  +  j(a2  +  62)  =  ci  +  jcz, 

where 

Ci  =  ai  +  bi  and  c2  =  a2  +  bz. 

Multiplication. — We  have,  evidently, 

Oi  +  ja2)(6i  +  J62)  =  ai&i  -  a2b2  +  j(a^  -f  biaz)  =  di  +  jfa, 

where 

di  =  dibi  —  a^bz',  dz  =  Q>ib%  +  b\a,z. 

In  general,  if  «i  +  jaz  =  bi  +  jb2  then  ai  =  61  and  a2  =  62- 
Power. — At  any  instant, 

p  =  ei, 

where  e  and  i  are  instantaneous  values  of  voltage  and  current. 
In  the  case  of  sine  waves,  where  e  =  Em  sin  ui  and  i  =  Im  sin 
(«*  +  «), 

p  =  ei  =  Emlm  sin  cot  sin  (co£  +  «)• 
8  113 


114 


ELECTRICAL  ENGINEERING 


Problem  49. — Plot  waves  of  voltage  and  current,  and  by  multiplying 
their  values  at  certain  instants  along  the  curves  show  that  the  resulting 
power  curve  is  a  sine  wave  of  double  frequency. 

Let  Em  =  1.4,  Im  =  0.7 

(1)  a  =  0°  (2)  a  =  45°  (3)  a  =  90°. 

Fig.  90  shows  the  curves  plotted  for  the  case  of  a.  =  0°.  The 
energy  developed  in  the  circuit,  in  any  time  dt  is  pdt.  The  total 

energy  during  a  cycle  is  then  J]T 
pdtj  where  T  is  the  time  of  one 
complete  cycle.     But  this  is  the 
-   area  enclosed  by  the  power  curve 
and  axis,  shown  shaded.     As  the 
values  of  power  are  always  posi- 
FIG.  90.  tive,    the    area   represents    energy 

expended,  or  work  done. 

The  student  should  show  that  when  a  is  not  0,  there  is  also 
negative  power,  which  represents  energy  returned  to  the  source, 
the  total  energy  expended  during  a  cycle  being  the  difference 
between  the  positive  and  negative  areas  enclosed  by  the  power 
curve  and  the  axis. 
Average  Value  of  Power  during  a  Period. -r-This  will  be, 


p 


mlm  sin  ut  sin  (coZ  -f  a)dt, 


which,  the  student  should  show,  is 


cos  a. 


This  may  be  written 


Em      I, 


V2  V2 


COS  a  =  El  cos  a 


(36) 


where  E  and  I  are  effective  values  as  usual. 

Thus  the  important  result  is  found,  that,  in 
case  of  sinusoidal  current  and  voltage  waves, 
the  average  power  is  equal  to  the  effective 
value  of  the  current  times  the  effective  value 
of  the  voltage  into  the  cosine  of  the  phase 
angle  between  the  two.  This  is  illustrated  in  Fig.  91,  and  it  is 
seen  that  when  7  is  zero  vector  =  i,  P  =  El  cos  a  =  el  =  ei. 
Similarly,  when  E  is  zero  vector  =  e,  P  =  Ei  =  ei. 


FIG.  91. 


METHOD  IN  TRANSMISSION  LINE  CALC  ULA  TION  1 15 


Power  is  obtained  by  multiplying  either  quantity  by  the  pro- 
jection of  the  other  upon  it. 

In  general,  if  E  makes  an  angle  7,  and  7  an  angle  0  with  the 
zero  axis, 

where  a  =  0  —  7, 

Therefore,  0>7. 

P  =  EI  cos  a  =  #7  cos  (0  -  7) 

=  EI  (cos  0  cos  7  +  sin  0  sin  7)  (36) 

t  e 


But  cos 


cos  7  == 


sin/3  =  —  sin  7  =  — 

Substituting  these  values  in 

P  =  ei  +  e'i' 

which  is  the  general  expression  for  the  average  power. 
Example. — Let  E  =  e  +  je' 

i  =  i  -f  jir. 

Then  by  (37)         P  =  ei  +  e'i'. 

Suppose,  however,  that  we  carry  out  the 
multiplication  of  the  vectors.     Thus, 

EI  =  (e+je')(i+jif) 

=  ei  —  e'i'  +  j(ei'  +  e'i) 

The  numerical  value  of  this  expression  is 


(36) 
(37) 


FIG.  92. 


\(«t-  eY)2  +  (e'i-  ei')2 

which  is  obviously  not  the  same  as  (37),  neither  is  its  real  com- 
ponent the  power,  since  it  has  a  minus  sign. 

It  has  been  shown  in  Fig.  90,  that  power  is  a  quantity  of  double 
frequency.  It  can  therefore  have  no  phase  relationship  with 
E  or  7.  Hence,  in  the  case  of  power  or  any  double  frequency 
quantity,  the  operation  of  multiplying  single  frequency  quanti- 
ties is  inadmissible. 

On  the  other  hand,  it  is  known  that  the  product 

(i-ji>)  (r+jx)  =  E 
is  quite  correct,  since  the  fundamental  frequency  only  is  involved. 


116  ELECTRICAL  ENGINEERING 

The  operation  of  obtaining  the  power  from  two  vectors 
E  and  I,  is  called  " telescoping"  the  vectors.  Thus,  the  prod- 
uct of  the  "real"  components  is  added  to  the  product  of  the 
"imaginary"  components,  without  any  change  of  sign  due  to 
the  presence  of  j. 

Power  Factor. — In  the  expression  for  power  (36)  the  term 
cos  a  is  called  the  power  factor.  The  product  El  represents  true 
power  only  in  certain  special  cases,  particularly  with  direct 
currents. 

T»  r  v     j  ^     j  .-         true  power 

Power  factor  may  be  defined  as  the  ratio  -  — ' 

apparent  power 

where  apparent  power  =  El. 

El  is  also  called  the  volt-amperes. 
Since  El  cos  a  is  the  true  power, 

.  El  cos  a 

power  factor  =  p.f.  =  — ^7 —  =  cos  a. 

Jiil 

Transmission  Line  Calculation. — The  calculation  of  circuits 
may  now  be  continued  to  include  the  case  which  represents  a 
simple  transmission  line  possessing  concentrated  resistance  and 
inductive  reactance,  being  supplied  with  power  at  one  end  by  a 

generator,  or  source  of  alternating 
current,  and  terminating  at  the  other 
end  in  any  prescribed  load.  A  cus- 
tomer usually  desires  constant  volt- 
FlG  93  age,  E,  at  the  load. 

In  Fig.   93  is  shown  a  generator 

supplying  power  over  a  transmission  line  of  impedance,  Z  =  r 
+  jxt  at  voltage  EQ  to  a  load,  the  current  of  which  is  i  +  ji' 
at  voltage  E.  E  and  i  are  in  time  phase ;  E  and  if  are  in  time 
quadrature. 

(1)  Let  E  be  known,  and  be  taken  as  the  zero  vector,  =  e. 
Then,  the  voltage  at  the  generator  terminals, 

Eo  =  e  +  IZ  =  e  +  (i  +  ji')  (r  +  jx) 
=  e  -f  ir  +  ijx  +  ji'r  —  i'x 
=  e  +  ir  —  i'x  +  j(ix  +  i'r)  =  a  -f-  jb, 
where 

a  =  e  -f  ir  —  i'x, 
b  =  ix  +  i'r. 


METHOD  IN  TRANSMISSION  LINE  CALCULATION  117 

The  power  factor  of  the  load,  cos  a  =  -=.  =  y 

Generator  volt-amp.  =  IEQ. 

Power  of  Generator. — P*Q  =  E0I  cos  7,  where  7  is  the  angle 
between  EQ  and  7. 

Vector  relationships  are  shown  in  Fig.  94:  (a),  for  leading 
and  (6),  for  lagging  current.     In  the  first  case 

7  =  a  —  |3  is  the  angle  between  EQ  and  7. 

/.  Cos  7  =  cos  (a  —  |8)  =  cos  a  cos  0  +  sin  a  sin  /3. 

i  i' 

But  cos  a  =  j,  and  sin  «  =  y-' 

a  6 

Likewise,  cos /5  =  yr  and  sin  0  =  ^- 

/.  Cos  7  =  yyr  (m  +  i'6). 

Substituting    this    value    into   the 
equation  for  power  of  generator, 

PQ    =    E0I  COS  7    =    itt  +   *'6. 

P0  could  be  more  quickly  obtained 
by  simply  telescoping  the  vectors  a 
+  jb  and  i  -f  ji'. 


Power  factor  at  the  generator 


power  PQ 


Ei 


Efficiency  of  transmission  =  ^5 

-T 


output 


Apparent  efficiency  ==  ratio, 
Regulation  =  --  ^  -- 


Having  obtained  the  general  expression  for  the  various  quan- 
tities which  enter  in,  we  may  now  take  a  specific  example  of 
transmission  line  calculation. 

Problem  60.—  In  Fig.  93,  let  E  =  1,  r  =  0.1,  x  =  0.2,  i  =  1. 

Let  i'  vary  from    —  1  to   +1. 

Determine  all  the  quantities,  i.e.,  current,  generator  voltage,  volt- 
amperes,  power  factor,  power,  transmission  efficiency,  apparent  efficiency 
and  regulation. 


118 


ELECTRICAL  ENGINEERING 


Tabulating: 


1 

j'  '. 

-1.0 

-0.75 

-0.5 

-0.25 

0.0 

0.25 

0.5 

0.75 

1.0 

If  

0.1 

0.1 

0.1 

0.1 

0.1 

0.1 

0.1 

0.1 

0.1 

—  i'x  

0.2 

0.15 

0.1 

0.05    * 

0.0 

-0.05 

-0.1 

-0.15 

-0.2 

a  

1.3 

1.25 

1.2 

1.15 

1.1 

1.05 

1.0 

0.95 

0.9 

ix  

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

0.2 

i'r  

-0.1 

—  0.075 

—  0.05 

—  0.025 

0.0 

0.025 

0.05 

0.075 

0.1 

I 

0.1 

0.  125 

0.  15 

0.  175 

0.2 

0.225 

0.25 

0.275 

0.3 

aj   . 

1.69 

1.56 

1  .44 

1.32 

1  .21 

1.10 

1.0 

0.90 

0.81 

62 

0.01 

0.0156 

0  .  0225 

0  .  0306 

0.04 

0  .  0506 

0.0625 

0.0756 

0.09 

a2  +  62  

1.7 

1  .  5756 

1.4625 

1.3506 

1.25 

1.1506 

1.0625 

0.9756 

0.9 

-y/02  4.  b2..  .  . 

1.31 

1.25 

1.21 

1.16 

.12 

1.08 

1.03 

0.985 

0.95 

Eo 

1.31 

1.25* 

1.21 

1.  16 

.12 

1.08 

1.03 

0.985 

0.95 

i'2 

1.0 

0.56 

0.25 

0.0625 

.0 

0  .  0625 

0.25 

0.56 

1.0 

i2  +  i'2 

2.0 

1.56 

1.25 

1  .0625 

.0 

1.0625 

1.25 

1.56 

2.0 

V*2  +  i'2..-. 

1.41 

1.25 

1.12 

1.03 

.0 

1.03 

1.12 

1.25 

1.41 

I  

1.41 

1.25 

1.12 

1.03 

.0 

1.03 

1.12 

1.25 

1.41 

E0I  

1.85 

1.56 

1.36 

1.23 

.12 

1.11 

1.15 

1.23 

1.34 

i/I  

0.707 

0.8 

0.895 

0.94 

.0 

0.94 

0.895 

0.8 

0.707 

CM 

1.3 

1.25 

1.2 

1.15 

.1 

1.05 

1.0 

0.95 

0.9 

bi'. 

—  0.1 

—  0.0937 

—  0.075 

—  0.0437 

0.0 

0.0563 

0.125 

0.206 

0.3 

Po  

1.2 

1.156 

1.125 

1.106 

1.1 

1.106 

1.125 

1.156 

1.2 

i'/i  

-1.0 

-0.75 

-0.5 

-0.25 

0.0 

0.25 

0.5 

0.75 

1.0 

tan  a  

-1.0 

-0.75 

-0.5 

-0.25 

0.0 

0.25 

0.5 

0.75 

1.0 

a  

45° 

37° 

26°  30' 

14° 

0° 

14° 

26°  30' 

37° 

45° 

Cos  a  

0.707 

0.8 

0.895 

0.94 

1.0 

0.94 

0.895 

0.8 

0.707 

P0/EoI  

0.65 

0.742 

0.827 

0.9 

0.982 

0.999 

0.978 

0.941 

0.895 

Cos  y  

0.65 

0.742 

0.827 

0.9 

0.982 

0.999 

0.978 

0.941 

0.895 

Fff         Ei 
•tin.  =  •p~"  ' 

0.834 

0.864 

0.89 

0.903 

0.907 

0.906 

0.89 

0.865 

0.835 

App.  eff.  = 

Ei 

0.54 

0.64 

0.735 

0.813 

0.894 

0.9 

0.87 

0.813 

0.74 

Eol 

Regulation  = 

Eo  -  E 

0.31 

0.25 

0.21 

0.16 

0.12 

0.08 

0.03 

-0.015 

-0.05 

E 

Problem  51. — Draw  vector  diagrams  for  the  cases  of  i'  —  —  1,  —  0.5, 
0,  0.5,  1,  of  problem  50  showing  the  relative  positions  of  E0,  E  and  /.  Also 
plot  the  curves  of  regulation  vs.  power  factor  of  generator  and  load. 

The  preceding  example,  like  many  others  included  in  this  book,  is  con- 
structed on  the  basis  of  percentages.  That  is,  by  choosing  e  =  1  and  i  =  1, 
whence  p  =  ei  =  1,  the  results  obtained  may  be  made  to  apply  to  any 
case  in  which  the  constants,  r  and  x,  give  the  same  percentage  of  ri  and  xi 


0.1  =  10  per  cent.,  ~  =  0.2  =  20  per  cent. 


drops.     In  this  example,  — 

If,  now,  10  per  cent,  resistance  drop  and  20  per  cent,  reactance  drop  be 
specified,  let  it  be  required  to  find,  with  varying  power  factor  of  the  load, 
the  same  quantities  determined  in  problem  50,  whenv  e  =  2200  volts  and 
i  =  300  amp. 

All  that  is  necessary,  now,  is  to  multiply  those  quantities  representing 
voltage  by  2200,  those  representing  current  by  300,  and  those  representing 
watts,  or  volt-amperes,  by  2200  X  300  =  660,000. 


METHOD  IN  TRANSMISSION  LINE  CALCULATION  119 

Thus,  for  i'  =  -  1  X  300  =  -  300,  E0  =  1.31  X  2200  =  2882,  /  = 
1.41  X  300  =  423,  E0I  =  1.85  X  660,000  =  1,221,000,  P0  =  1.2  X  660,- 
000  =  792,000. 

The  other  quantities  sought — power  factor  of  load  and  of  generator, 
efficiency,  apparent  efficiency  and  regulation — are  the  same  as  already 
calculated. 

The  advantages  of  problems  on  the  percentage  basis  are  thus  quite 
obvious. 

Problem  62. — A  transmission  line  1  mile  long  supplies  power  to  a  load  of 
100  kw.  and  1000  volts  at  power  factor  of  0.8  and  frequency  of  60  cycles. 

The  line  is  composed  of  two  parallel  No.  000  B.  &  S.  wires,  18  in.  apart. 

Find  generator  voltage,  current,  power  factor,  power  output,  line  effi- 
ciency, apparent  efficiency,  regulation,  with  the  current  both  lagging  and 
leading. 

The  resistance  of  No.  000  B.  &  S.  hard-drawn  copper  wire  may  be  taken 
as  0.06  ohm  per  1000  ft.  at  20°C. 

The  reactance  is  2wfL,  where  L  is  the  inductance,  in  henrys,  per  centi- 
meter length  of  wire.  L  may  be  calculated  from  the  formula, 


109 

where  D  and  r  are,  respectively,  the  distance  between 
centers  of  wires  and  the  radius  of  the  wire  (Fig.  95). 


D 
FIG.  95 


CHAPTER  XVII 

CONSTANT  POTENTIAL-CONSTANT  CURRENT 
TRANSFORMATION 

It  is  sometimes  desirable  that  the  current  in  a  circuit  shall 
remain  constant  while  the  load  varies.  In  series  lighting  circuits, 
for  example,  the  current  through  each  lamp  must  be  nearly 
constant,  while  the  number  of  lamps  may  vary  from  none  at 
all  up  to  the  most  that  the  system  can  sustain.  Generally, 
however,  it  is  desirable  that  the  energy  shall  be  supplied  from 
a  source  of  constant  potential,  such  as  a  constant  potential 
generator.  Such  a  system  is  possible  with  a  circuit  arrangement 
like  that  shown  in  Fig.  96.  Here,  a  high  resistance,  r,  is  placed 
in  series  with  the  lamps.  When  the  lamps  are  comparatively 
few,  changing  their  number  will  not  alter  the  total  resistance  of 
the  circuit  very  much,  and  the  current  will  therefore  be  fairly 
constant. 

This  arrangement  is  not,  however,  economical,  as  a  large  pro- 
portion of  the  power  developed  is  always  lost  in  the  resistance,  r. 


FIG.  96.  FIG.  97. 

We  may,  therefore,  try  substituting  inductive  reactance,  x,  for 
r,  and  determine  if  this  will  give  better  results. 

In  this  case,  Fig.  97,  let  the  generator  voltage  be  unity,  that 
is  e  =  1.  Let  the  largest  value  of  the  permissible  current,  in 
the  circuit  also  be  unity,  that  is,  /  =  1,  and  let  the  resistance  of 
the  lamps,  that  is,  the  number  of  the  lamps,  vary.  We  may 
then  find  the  maximum  resistance  of  the  lamps  which  may  be 
obtained  without  reducing  the  current  lower  than,  say,  0.925, 
which  will  be  considered  the  minimum,  permissible  current. 
The  current  is  obviously  a  maximum  when  the  resistance  is 
zero,  that  is,  when  no  lamps  are  used. 

e       I 
Then,  x  =  j  =  j  =  1  ohm. 

120 


CONSTANT  CURRENT  TRANSFORMATION       121 


Let  r  be  the  resistance  of  the  lamps,    r  is  variable,  depending  on 
the  number  of  lamps  in  circuit  at  any  time. 
Let  E  be  made  the  zero  vector,  =  e 

e  e  e  (r  -  jx) 


Then 


where 


I  = 


7  =  e 
Tabulating  for  varying  r: 


(38) 


r 

0 

0.1 

0.2 

0.4 

0.6 

0.8 

1 

r2  +  X2. 

1 

1  01 

1  04 

1  16 

1  36 

1  64 

2  0 

a  

0 

0.099 

0.192 

0.345 

0.441 

0.488 

0.5 

b  

-1 

-0  99 

-0  96 

—0  861 

-0  735 

-0  61 

-0  5 

a2 

o 

0  0098 

0  0369 

0  119 

0  195 

0  238 

0  25 

62  

1 

0.98 

0.921 

0.741 

0.54 

0.373 

0.25 

a2  +  62.. 

1 

0  9898 

0  9579 

0  86 

0  735 

0  611 

0.5 

Va2  +  62  
7  

1 
1 

0.994 
0.994 

0.978 
0.978 

0.927 
0.927 

0.857 
0.857 

0.782 
0.782 

0.707 
0.707 

It  is  evident  from  the  calculation  that  the  limit  of  current 
is  reached  by  a  resistance  of  0.4  ohm.  This  resistance  could 
evidently  be  obtained  by  directly  substituting  the  value  /  =  0.925 
into  Eq.  (38)  and  solving  for  r.  However,  it  is  frequently  prefer- 
able to  carry  out  the  tabulation,  thus  gaining  the  material  for 
plotting  the  curves.  These  curves  are  far  more  instructive  than 
the  mere  numerical  answer. 

In  this  case,  where  reactance  has  been  used  instead  of  resist- 
ance in  order  to  obtain  (approx.)  constant  current,  all  the 
energy  is  consumed  in  the  lamps  themselves  since  the  reactance 
(assuming  zero  resistance)  does  not  consume  any  energy.  Thus 
the  system  is  efficient.  However,  the  power  factor  appears  to  be 
very  low. 

Problem  63. — Determine  the  power  factor  of  the  circuit  and  plot  it 
against  the  resistance  of  the  lamps. 

Altogether,  it  may  be  said  that  in  practice  this  arrangement  is  cheap  and 
practical.  A  constant-current  "tub"  transformer  has  a  higher  power 
factor,  but  is  also  more  expensive.  In  this  machine  regulation  is  obtained 
by  altering  the  reactance  in  the  circuit  by  means  of  the  repulsion  between 
the  primary  and  the  secondary  coils. 

Problem  64. — A  constant-current  system  is  supplied  with  power  by  a 


122 


ELECTRICAL  ENGINEERING 


generator  at  2300  volts  and  60  cycles.  The  line  resistance  is  negligible. 
Each  lamp  has  6  ohms  resistance.  Current  must  be  maintained  between 
the  limits  of  7.2  and  6  amp. 

Find  the  maximum  number  of  lamps,  both  by  the  "resistance"  and  by 
the  "reactance"  method  of  obtaining  constant  current.  Plot  and  compare 
curves  of  number  of  lamps  vs.  current  for  the  2  cases. 

Solution. — (a)  Resistance  method. 
Let  r  =  res.  in  series,  TL  =  res.  of  lamps. 


_  2300 

Then  r  =  -=-^ 


r  +  -TL  = 

rL  = 

No.  lamps  = 


2300 


=  320  ohms. 

=  383.3  ohms. 

63.3  ohms. 


383.3  -  320 

~   =  10.5  =  10  lamps. 


Lamps 

0 

2 

4 

6 

8 

10 

12 

TL 

0  0 

12  0 

24  0 

36  0 

48  0 

60  0 

72  0 

r  +  rL.. 

320.0 

332.0 

344.0 

356.0 

368.0 

380.0 

392.0 

I  

7.2 

6.93 

6.68 

6.46 

6.25 

6.05 

5.87 

(6)  Reactance  method.     Neglecting  the  resistance  of  the  reactance  coil, 

2300 


rL2  +  3202 
No.  lamps 


2300 
6 

383.3 
210.8 


=  320  ohms. 
=  383.3  ohms. 
/.  TL  =  210.8 
35.1  =  35  lamps. 


Lamps 

0 

10 

20 

30 

35 

40 

TL  
r,2  . 

0 

A 

60 

QfiOO 

120 
14  400 

180 

Q9  CJOO 

210 
44  200 

240 
K7  700 

z2- 

102,500 

102,500 

102,500 

102,500 

102,500 

102,500 

rL*+x*  

102,500 

106,100 

116,900 

135,000 

146,700 

160,200 

W+z2  
/  

320 

7.2 

326 
7.05 

342 
6.72 

369 
6.23 

383 
6.00 

400 
5.75 

NOTE. — In  an  example  of  this  kind  it  may  be  as  convenient  to  work 
directly,  without  the  use  of  complex  quantities  as  has  just  been  done. 
With  more  complicated  circuits,  however,  it  may  be  far  more  convenient 
and  far  safer  to  adhere  strictly  to  the  complex  method. 

There  are  many  other  schemes  for  obtaining  constant  current. 
Some  of  these  involve  the  use  of  condensers  which  are  treated 
in  the  next  chapter. 

This  subject  will  be  discussed  further  in  Chap.  XXI. 


CHAPTER  XVIII 
CAPACITY  AND  CAPACITY  REACTANCE 

Two  conducting  surfaces,  insulated  from  each  other,  are  said 
to  possess  electro-static  capacity.  Such  an  arrangement  em- 
bodied as  a  piece  of  apparatus  is  called  an  electrical  condenser. 

Condenser. — When  the  plates  of  a  condenser  are  connected 
respectively  to  the  positive  and  negative  terminals  of  a  direct- 
current  generator,  the  condenser  becomes  charged.  That  is, 
when  a  switch,  s,  Fig.  98,  is  closed,  completing  the  circuit  con- 
taining the  generator  and  the  condenser,  ammeters  A,  placed  in  the 
leads,  will  indicate  a  momentary  current  in 
the  direction  of  the  arrows.  No  current,  in 
the  ordinary  sense,  could  pass  between  the  t 
plates.  The  phenomenon  thus  resembles 
the  piling  up  of  electricity,  as  so  much  ma-  F 

terial,  on  one  plate,  the  positive  plate,  since 
it  is  connected  to  the  positive  terminal  of  the  generator,  and 
the  withdrawing  of  an  equal  amount  of  electricity  from  the  other, 
the  negative  plate.  This  quantity  of  electricity  which  seems  to 
have  been  transferred  from  one  plate  to  the  other  is  the  charge 
placed  upon  the  condenser.  The  condenser  is  maintained  in  an 
unstable  state  by  the  e.m.f.  of  the  generator.  If  the  generator 
is  disconnected,  the  condenser  continues  to  remain  charged  so 
long  as  its  plates  remain  insulated  from  each  other,  but  as  soon  as 
electrical  connection  is  made  between  them,  the  condenser  dis- 
charges itself  by  a  rush  of  electricity  from  the  positive  to  the 
negative  plate,  as  indicated  by  the  flow  of  electricity  or  current 
through  the  meters.  If  the  condenser  is  charged  to  a  difference 
of  potential  which  is  excessive,  the  insulating  dielectric  breaks 
down,  allowing  a  discharge  to  take  place  between  the  plates. 
This  indicates  that  the  dielectric  is  placed  under  a  strain  when  the 
condenser  is  charged.  In  fact,  the  dielectric  behaves  much  like 
an  elastic  medium  compressed  between  plates.  When  the 
pressure  is  removed  the  medium  assumes  its  normal  condition. 

The  plates  act  merely  as  carriers  or  distributors  of  the  charge, 

123 


124  ELECTRICAL  ENGINEERING 

while  its  actual  seat,  as  found  out  by  FRANKLIN,  is  the  surface  of 
the  dielectric. 

The  capacity  of  any  given  condenser  is  determined  by  the 
dimensions  of  its  plates,  their  distance  apart,  and  the  nature  of 
the  dielectric  which  separates  them. 

Capacity  is  not  a  property  solely  of  apparatus  arranged  in  the 
form  of  a  condenser,  but  any  body  may  be  said  to  possess 
capacity  —  for  instance,  a  metallic  sphere,  insulated  and  isolated 
in  space.  But  this  may  also  be  considered  as  a  limiting  form 
of  condenser  in  which  one  plate  is  the  surface  of  the  sphere  and 
the  other  is  a  surrounding  sphere  of  infinite  radius.  In  this 
case  the  strain  in  the  dielectric  may  be  represented  by  the  lines  of 
force,  or  tubes  of  force,  extending  radially  outward  from  the 
surface  of  the  sphere  and  terminating  on  the  surface  of  the 
imaginary  sphere  infinitely  distant.  This  conception  of  lines, 
or  tubes  of  force,  due  to  FARADAY,  makes  the  direction  of  a  line 
or  axis  of  a  tube  the  direction  of  the  force  at  any  point,  and  the 
number  per  square  centimeter,  or  density,  of  lines  or  tubes  be- 
comes a  measure  of  the  force  at  the  point. 

FARADAY  assumed  that  the  number  of  tubes  is  the  same 
numerically  as  the  charge  per  unit  surface,  and  that  the  number 
of  lines  emanating  from  a  charge  Q  is  $  =  4?rQ.  Thus  each  tube 
contains  4ir  lines  of  force. 

By  connecting  a  sphere  to  one  terminal  of  a  battery,  Fig.  99, 
and  connecting  the  other  terminal  to  earth,  assumed  infinitely 
distant,  we  establish  a  number  of  electric  lines 
/      of  force  extending  outward  from  the  sphere. 
The  number  of  lines  established  is  471-Q,1  where 
Q  is  the  amount  of  the  charge  placed  upon  the 
sphere. 

If  the  sphere  is  in  air,  the  practical  limit  to 
^FIG   99  ^ne   num^er   °f  nnes   which  it  is  possible  to 

establish  is  very  closely  100  per  sq.  cm.  of  sur- 
face.    Thus,  to  produce  100  lines  per  sq.  cm.  requires  a  charge 

100 
Q  =  -7—  =  8  absolute  electro-static  units  per  sq.  cm. 

To  increase  the  number  of  lines  established  in  any  given 
case,  the  difference  of  potential,  or  voltage,  should  be  increased. 
These  lines  are  conceived  to  displace  the  ether,  until  by  continu- 
ally increasing  the  voltage,  the  crowding  of  them  becomes  so 

1  See  Advanced  Course  in  Electrical  Engineering. 


\\  I  / 
" 


CAPACITY  AND  CAPACITY  REACTANCE       125 

great  that  the  dielectric  breaks  down.  The  ability  of  any  dielec- 
tric to  withstand  rupture  under  the  strain  of  potential-difference 
is  called  "dielectric  strength." 

With  a  parallel  plate  condenser,  Fig.     -f~ 
100,   the  lines  or  tubes  are  parallel,     zip 
except  at  the  edges,  where  they  bow    z=jT 
outward. 

By   definition,    the*  charge  due  to 
current  i  during  interval  dt  is: 

dq  =  idt.  (39) 

The  practical  unit  of  charge  or  quantity,  q,  is  the  coulomb. 
Another  fundamental  relation  is  that 

q  =  Ce  (40) 

where  C  is  the  capacity,  and  e  the  e.m.f.  or  difference  of  potential. 
This  law,  found  experimentally,  shows  that  the  number  of  tubes 
which  can  be  set  up  in  a  condenser  of  capacity  C  depends  directly 
on  the  potential  difference.  In  practical  units,  the  charge  in 
coulombs  is  equal  to  the  product  of  the  capacity  in  farads  and  the 
potential  difference  in  volts.  "Charge"  is  not  a  material  quan- 
tity, but  may  well  be  thought  of  as  a  measure  of  "tubes." 
Substituting  from  (39)  into  (40),  since 

*-•*, 

and 

dq  =  Cde, 


which  is  called  the  charging  current,  or  capacity  current  of  the 
condenser. 

Assuming  a  sine  wave  of  e.m.f.,  impressed  on  a  condenser 

then,  e  =  EM  sin  wt  (42) 

i  =  CEMw  cos  orf.  (43) 

The  capacity,  C,  is  a  constant  of  the  circuit,  that  is,  like  resist- 
ance and  inductance,  it  is  a  quantity  fixed  by  the  mechanical 
arrangement  of  the  circuit. 

Eq.  (42)  may  be  written: 

i  =  CEmo>  sin  (cd  +  90°).  (43a) 


126  ELECTRICAL  ENGINEERING 

Comparing  (42)  and  (43a)  it  is  seen  that  the  charging  current 
is  90°  in  time  phase  ahead  of  e.     Also, 


The  effective  value  of  the  charging  current  is  then 


whence 

E 


Xc. 


/  27T/C 

The  quantity  xc  is  called  capacity  reactance,  and  its  use  in  cir- 
cuit calculations  is  similar  to  inductive  reactance. 

Expression  of  Condensive  Impedance. — It  has  been  shown  that 
the  charging  current  leads  the  impressed  e.m.f.  90°  in  time. 

Thus,  if  the  charging  current  /  is  made  zero  vector,  the  im- 
pressed e.m.f.  is  —  jkl  where  k  is  some  constant  and  is  obviously 
xc.  Thus  E  =  —  jxj. 

In  an  inductive  circuit  the  current  lags  90°  behind  the  im- 
pressed e.m.f. 

Thus  E  =  jxl. 

Convention  has  settled  that  an  inductive  impedance  is  Z  = 
r  +  jx\  thus  the  condensive  impedance  is  Z  =  r  —  jxc  where  xc 

as  well  as  x  is  always  a  positive  number. 

Circuit  Containing  Resistance,  Inductance  and  Capacity  in 
Series. — To  find  the  current.  Let  E,  the  impressed  e.m.f.,  be 
the  zero  vector.  Then 

/  =  -     -A-     -  = 
r  +  jx  -  jxc 

where 


a  = 


b  =  - 


^X         Xc) 

(x  -  xc) 


r  (x  -  zc)»- 
To  find  the  voltage  Ec  across  the  condenser.     We  have: 

EC     =     I(-jXc) 

=  e(a  H-  jb)(—  jxc)  =  e(—  ajxc  +  bxc). 
Similarly,  the  voltage  across  the  inductance  is 
EL  =  I  X  jx 

=  e(a  +  jb)jx  =  e(ajx  —  bx). 


CAPACITY  AND  CAPACITY  REACTANCE        127 


Problem  56.  —  Let  the  constants  of  a  circuit  be  r  =  1  ohm,  L  =  0.0265 
henry,  C  =  0.000265  farad,  and  let  100  volts  be  impressed  on  the  circuit 
at  variable  frequency.     Find,  and  plot  against 
the  frequency,  7,  Ec,  EL,  Er  for  frequencies  from 
0  to  100  cycles  per  sec.  Jl  »/.     i       fo 

Solution.  —  We  have: 
7 


*«np-  f c 


e(a  +  jb);  I  =  eV  a2  +  b2. 
Ec  =  e(  -  ajxc  +  bxc);  Ec  = 
EL  =  e(ajx  -  bx);  EL  = 
Also, 

ET  =  e(a  +  j'6)r;  Er  =  er\/a2 
Tabulating: 


FIG.  102. 


/ 

0 

20 

40 

50 

55 

60 

65 

70 

100 

2irf 

0 

125.6 

251.2 

314.0 

345.2 

376.8 

408.0 

440.0 

628.0 

X 

0 

3.33 

5.65 

8.33 

9.15*' 

10.0 

10.8 

11.65 

16.66 

xc 

30.0 

17.65 

12.0 

10.92 

10.0 

9.25 

8.57 

6.0 

(x  -  xc)  

-     CO 

-26.67 

-12.0 

-3.67 

-1.77 

0.0 

1.55 

3.08 

10.66 

(X   -   XC)2.  .  .  . 

+   oo2 

712.0 

144.0 

13.5 

3.14 

0.0 

2.4 

9.5 

114.0 

r2  +  (x-xc)2.. 

oo2 

713.0 

145.0 

14.5 

4.14 

1.0 

3.4 

10.5 

115.0 

a   

0 

0.0014 

0.0069 

0.069 

0.242 

1.0 

0.294 

0.095 

0.0087 

b  

0 

0.0374 

0.0827 

0.253 

0.428 

0.0 

L0.455 

-0.293 

-0.0925 

a*  

0 

0.000002 

0  .  000048 

0.0048 

0.059 

1.0 

0.086 

0.009 

0.000076 

62  

0 

0.0014 

0.0068 

0.064 

0.183 

0.0 

0.207 

0.086 

0  .  0086 

(I*    +   62  

0 

0.0014 

0.0069 

0.0688 

0.242 

1.0 

0.293 

0.095 

0  .  0087 

Va2  +  62.  ... 

0 

0.0374 

0.0828 

0.262 

0.492 

1.0 

0.54 

0.308 

0.093 

7  

0 

3.74 

8.28 

26.2 

49.2 

100.0 

54.0 

30.8 

9.3 

EC  

0 

112.2 

146.1 

314.0 

537.0 

1000.0 

500.0 

264.0 

55.8 

EL  

0 

12.5 

46.8 

218.0 

450.0 

1000  .  0 

583.0 

370.0 

155.0 

Er  

0 

3.74 

8.28 

26.2 

49.2 

100.0 

54.0 

30.8 

9.3 

40  60 

Frequency 
FIG.  103. 


128  ELECTRICAL  ENGINEERING 

Resonance. — Curves  of  the  form  shown  in  Fig.  103  are 
called  resonance  curves,  and  their  maximum  points  of  the  de- 
pendent variables  are  called  resonance  points.  In  this  case,  it 
is  said  that  60  cycles  is  the  frequency  of  resonance. 

On  examining  the  problem  it  is  seen  that  resonance  is  attained 
at  that  frequency  for  which  x  —  xc  =  0,  or  when  the  effect  of 
inductance  is  just  nullified  by  that  of  capacity.  The  circuit  then 
behaves  as  though  it  possessed  resistance  only. 


CHAPTER  XIX 
PARALLEL  CIRCUITS 

Let  1 1  and  1 2  be  any  currents  in  the  branches  of  a  parallel 
system,  such  that  I\  =  i\  +  jif\  and  72  =  it  +  jifz- 

Laying  off  these  vectors  (Fig.  105),  and  adding  them,  gives 
7  =  7i  +  7a  =  ii  +  la  +  j(i'i  +  *'2).  (44) 

Let  the  impedances  of  the  branches  (Fig.  104)  be  Zi  =  ri  —  jxi 
and  Z2  =  r2  +  j^2,  respectively. 


FIG.  104.  FIG.  105. 

To  find  the  currents  7i,  72,  and  7.     We  have: 

. —  _  e 


where  g\  = 


r\ 


+ 


T 


+  xi2 
is  the  conductance, 
is  the  susceptance, 


and  FI  =  gi  +  jbi  is  the  admittance  of  the  first  branch  circuit. 
eg i  is  the  power  component  of  I\. 
cbi  is  the  wattless  component  of  I\. 
Similarly, 

_£.  ^2    —  JXz)  ,  .        .,     v     _        y 

where  _^^ 


r»J 


and 


r2 


/= 


(45) 


129 


130  ELECTRICAL  ENGINEERING 

To  find  the  joint  impedance,  Z,  of  the  branches, 

e_       _e_        !_ 

~~  I    '  eY  ''=  Y' 

Example. — In  Fig.  104  let  n  =  rz  =  0,   and  Xi  =  ^-f^>  x-2  =  2ir/L. 
Then 

«•- 

=  0, 
1 


— 

"  z2  27T/L' 

Then,  from  (45), 

7  =   e  (o  +  j(2*/C  -  g^)  )  •  (46) 

From  this  it  is  seen  that  the  line  current  is  in  time  quadrature  with  the 
voltage. 

If  I  =  0,  then  from  (46)  we  have  the  relation 


or 

=    1. 


that  is  to  say,  that  if  in  a  circuit  such  as  is  here  considered  the  frequency 
be  varied,  a  value  may  be  reached  for  which  the  line  current  will  be  reduced 
to  zero.  In  such  a  case  the  currents  in  the  branches  will  be 

7i  =  e(0 


Both  of  these  currents  are  in  time  quadrature  with  the  voltage,  but 
/i  is  leading  while  72  is  lagging. 

Thus,  they  are  in  time  phase  opposition  to  each  other. 

Problem  66.—  In  the  circuit  of  Fig.  104  let  e  =  100,  n  =  r2  =  1,  L  = 
0.0265,  C  =  0.000265.  Let  the  frequency  vary,  as  in  problem  54.  Find 
/»  Ii,  Jzj  and  plot  them  against  the  frequency. 

Transmission  Line  Supplying  Power  to  Parallel  Loads.  —  Let 

a  transmission  line  of  impedance  ZQ  =  rQ  +  j%o  be  used  to  supply 
power  to  a  load  consisting  of  two  impedances,  Zi  =  7*1  +  jx\  and 
£2  =  r2  +  jxz,  which  are  in  parallel.  Besides  the  impedances, 
let  E  the  voltage  at  the  receiving  end  be  known. 

Find  7,  /i,  72,  EQ,  P.F.  of  generator  and  of  combined  load, 
regulation  and  efficiency  of  the  line. 

E  is  chosen  as  the  zero  vector  =  e. 


PARALLEL  CIRCUITS 


131 


Then 


where 


Zi      ri+jxi        rf  +  xi' 
«(0i  +  jbi),. 


Similarly, 


where 


/2    = 


FIG.  106. 


And 

where 
Then 


/    =  /I  +  /2 

=  m  +  jn, 


=  e(0i  4-  02  +  j(&i  +  62) 

\       __  _/t     i^  t  \ 

=  e  +  (w  -f  jn)(r0  +  jx0) 
—  nxQ  +  j(nro  - 


Eo  =  e  + 

=  e  + 

=  a0 
where 

aQ  =  e  -{-  mr0  —  nx0)  60  =  nr0 
Power  of  generator,  by  telescoping  EQ  and  I  =PQ  =  a0m  +  60n. 

PO         a0w  -f-  60?i 


.'.  P.F.  of  generator  = 
P.F.  of  combined  load  =  -v  =  — F' 


EJ 


-  e 


Regulation  = 


„„,  .  P  em 

Efficiency  =  ^-  = 


Problem  67. — In  the  same  circuit  (Fig.  106),  let  E0  be  known  and  E 
unknown.  Find  all  the  quantities  obtained  in  the  last  problem. 

NOTE. — In  solving  this  problem  the  student  is  again  urged  to  pay 
particular  attention  to  the  form  of  his  work.  In  order  to  add  emphasis  to 
this  matter  these  similar  problems  are  here  given,  the  one  being  worked  out 
and  the  other  left  for  the  student  to  do. 


132  ELECTRICAL  ENGINEERING 

The  numerical  or  scalar  expressions  are  not  put  down.  It  is  assumed 
that  they  may  always  be  obtained  when  needed  by  the  simple  process  of 
rationalizing  a  simple  complex  expression.  By  omitting  them  in  the  process, 
confusion  is  eliminated. 

Approximate  Transmission  Line  Calculation. — The  two  parallel 
wires  of  a  transmission  line  may  be  regarded  as  constituting  the 
plates  of  a  condenser.  When  alternating  e.m.f.  is  impressed 
upon  the  line  there  will  therefore  flow  a  charging  or  capacity 
current  over  the  line,  whether  the  distant  end  is  open  or  closed. 
Fig.  107  gives  an  approximate  representation  of  such  a  line  in 


___ 

' 


FIG.  107. 

which  the  line  capacity  is  replaced  by  two  condensers,  one  at 
each  end,  so  proportioned  that  each  shall  take  one-half  of  the 
charging  current.  The  charging  current  is  taken  as  2i2.  ii  is 
always  positive,  whereas  i',  the  wattless  component  of  the  load 
current,  is  positive  or  negative  depending  on  the  load. 
Then 


EQ  =  e  +  I  Z0  =  o0+  j 
where  ao  =  e  +  irQ  —  isxQ) 


/o  =  /  +  jit  =  i  +  j(i'  +  2i2)  =  i  +  ju. 

From  these,  the  power,  power  factor,  efficiency,  etc.,  may  be 
determined.  Expressions  should  be  obtained  by  the  student  for 
practice,  as  follows: 

Power  given  by  generator      =  P0  = 

Apparent  power  at  generator  =  EQIQ  = 

p 
P.F.  at  generator  =  cos  <*o       =  jrj~  — 

Efficiency  of  transmission        =  ,5-  = 

*  o 

Apparent  efficiency 


CHAPTER  XX 


DISTORTED  WAVES.     RESONANCE  EFFECTS 

So  far,  only  current  and  voltage  waves  have  been  dealt  with 
which  followed  a  sinusoidal  variation  with  respect  to  the  time 
and  had  the  same  frequency  or  period.  In  the  laboratory,  re- 
sults obtained  are  found  not  always  to  agree  with  those  expected 
from  the  theory.  This  is  frequently  due  to  the  assumption  in 
theory  of  pure  sine  waves,  whereas,  in  practice,  a  pure  sine  wave 
is  only  approximately  attainable,  and  the  actual  waves  may 
differ  greatly  from  that  form. 

It  can  be  proven  that  any  curve  representing  changes  occurring 
with  time  can  be  resolved  into  a  number  of  sine  waves  of  differ- 
ent frequency  —  as  long  as  the  curve  representing  the  changes 
is  a  univalent  function  of  time  —  which  it  always  is  in  electrical 
problems. 

It  can  also  be  proven  that  if  the  curves  traced  are  symmetrical 
above  and  below  the  axis  —  no  matter  how  distorted  —  the  sine 
waves  contain  only  the  odd  frequencies. 

Thus  assume  as  the  simplest  case  that  the  current  is  distorted 
in  such  a  way  that  it  can  be  •  represented  by  the  first  two  terms 
of  the  series,  that  is  that: 


i  =  Iim  sin  wt  -f  J3m  sin 


a). 


It  is  seen  that  the  frequency  of  the  second  component  wave 
is  three  times  that  of  the  first. 
The  first  wave  is  called  the 
fundamental  of  the  complex 
wave,  the  second  wave  is 
called  the  third  harmonic. 
The  angle  a  denotes  the  per- 
manent phase  difference  be- 
tween the  waves.  Such  a 
combination  of  waves  is  seen  ina 

rIG.    luo. 

to   be   a  distorted   wave,   as 

shown  in  Fig.  108.  To  find  the  amount  of  heat  such  a  wave 
will  develop  in  a  circuit,  that  is,  to  find  the  effective  value  of 
the  complex  wave. 

133 


134  ELECTRICAL  ENGINEERING 

Evidently  the  heat  developed  at  any  instant  is  proportional  to 
i2,    and 

i2  =  [Iim  sin  ut  +  73m  sin  (3co£  -f  a)]2. 


The  mean  value  of  the  heat  developed  during  a  cycle  of  the 
wave  will  then  be  proportional  to 

1   CT 
mean  i2  =  —  I  [Ilm  sin  ut  +  hm  sin  (3co£  +  a)]2dt.      (47) 

Thus,  the  effective  value  of  the  current  is 


\/mean^2  =  7  =  \/™  I  t^im  sin  cot  +  73w(3co£  +  «)]2cfr.  (48) 


The  student  should  solve  (47)  and  (48),  and  show  that 

1       2  7       2 

.«          -Mm       ,     -*  3m 

mean  z2       T^+jT  (49) 

and  /  =  \/Ii2  +  /a2 

where   7im  =  maximum  value  of  the  fundamental  current  wave, 
73m  =  maximum  value  of  the  third  harmonic, 

7i  =  -^7=  =  effective  value  of  the  fundamental, 


7am 

7s  =  — j=  —  effective  value  of  the  third  harmonic. 
V2 

Also,  in  general,  where  there  are  any  number  of  component 
harmonic  waves  in  a  circuit, 


/  =  Vli2  +  h2  +  762  +  .    .    .  (50) 

Thus  is  found  the  important  rule  that  the  effective  value 
(ammeter  reading)  of  any  number  of  currents  of  different  fre- 
quencies is  equal  to  the  square  root 
w  of  the  sum  of  the  squares  of  the  in- 

dividual effective  values. 

15 

NOTE. — Eq.  (50)  holds  for  any  combina- 
tion of  harmonics  whatsoever.  With  alter- 
nating-current machinery,  we  have  to  deal 


„  only   with  odd  harmonics,  as  the  positive 

and  negative  waves  are  always  symmetrical 
except  during  transient  periods  not  considered  in  this  volume. 

Example. — In  Fig.  109  are  represented  three  generators  which 
supply  respectively  20  amp.  at  60  cycles,  15  amp.  at  25  cycles,  and 
10  amp.  at  10  cycles.  They  all  use  a  common  wire  for  a  part 


DISTORTED  WAVES.     RESONANCE  EFFECTS    135 

of  their  circuits.     Then  the  current  which  flows  in  the  common 
wire  is 

/  =  V202  +  152  +  I62  =  27  amp. 

Problem  68.  —  If  still  another  generator  is  added  to  the  above  system  and 
it  supplies  12  amp.,  direct  current,  to  its  load,  using  the  common  wire,  find 
the  current,  I,  that  will  then  be  in  the  common  wire  and  explain  the  result. 

E.m.f.  Which  Causes  Distorted  Waves  of  Current.  —  If  the 

current  in  any  circuit  is  given  by  the  equation 

i  =  1  1  sin  at  +  7  3  sin  (3orf  +  a)  (51)' 

the  question  may  naturally  arise  as  to  what  kind  of  e.m.f.  wave 
will  cause  such  a  current  to  flow.    Will  the  e.m.f.  wave  be  more 
or  less  distorted  when  the  current  is  supplied  to  a  circuit  of  re- 
sistance and  inductive  reactance? 
We  have  (Eq.  15), 

,   T  di 

e  =  ir+Ldt' 
Substituting  from  (51), 

e  =  Ijr  sin  wt  +  73r  sin  (3co£  +  a)  +  L(/io>  cos  co£  + 

373co  cos  (3co£  +  a)) 
=  7  if  sin  ut  +  L/io)  cos  coZ  +  73r  sin  (3o>Z  +  a)  + 

3L73o>  cos  (3coZ  -f  a) 
—  lir  sin  ut  +  IiX  cos  ut  +  Isr  sin  (3orf  +  a)  + 

I3x3  cos  (3orf  +  a).  (52) 


Let          -  =  tan  /?;  —  =  tan  /33. 

Then 

r  =  Zi  cos  j8  =  23  cos  /33. 

Substituting  these  values  in  (52), 

e  =  /i£i(cos  j8  sin  co£  +  sin  0  cos  «$)•+  /3^s(cos  03  sin 

[3orf  +  a]  +  sin  &  cos  [3orf  +  a]) 
=  JiZi  sin  (coi  +  0)  +  73Za  sin  (3w<  +  a  +  ft) 
=  Ei  sin  (co£  +  0)  +  Ei  sin  (3arf  +  «  +  ft)-  (53) 


Thus  the  amplitude,  EI,  of  the  fundamental  voltage  wave  is 
Zi  times  that  of  the  current  fundamental;  the  amplitude  E9  of 
the  triple  frequency  voltage  wave  is  Z3  times  that  of  the  cur- 
rent triple  frequency  harmonic. 


136  ELECTRICAL  ENGINEERING 

The  difference  between  the  multipliers,  Zi  and  Z3,  is  due  to 
their  respective  reactances,  x  and  x3,  since  r  is  the  same  in  each. 
But  z3  is  3x. 

Therefore,  it  is  seen  that  the  triple  frequency  voltage  wave 
is  greater  in  proportion  to  its  fundamental  than  the  triple  fre- 
quency current  wave  to  its  fundamental.  In  other  words,  the 
voltage  wave  is  more  distorted. 

Conversely,  it  may  be  said  that  when  a  distorted  voltage  is 
impressed  on  a  circuit,  the  effect  of  the  inductive  reactance  is 
to  smooth  out  some  of  the  distortion  in  the  current  wave. 

Problem  69. — Show  that  when  the  e.m.f., 

e  =  EI  sin  (at  +  E3  sin  (at  +  a), 

is  impressed  on  a  circuit  of  resistance-  only,  the  current  flowing  will  have 
the  same  amount  of  distortion  as  the  voltage  has. 

Problem  60. — Show  that  when  the  e.m.f.  of  problem  59  is  impressed  on 
a  circuit  containing  resistance  and  capacity,  the  effect  of  the  capacity  is  to 
increase  the  distortion  of  the  current. 

If  the  voltage  (53)  is  measured  by  a  voltmeter,  what  will  the 
reading  be?  From  the  development  of  (51)  in  respect  to  dis- 
torted currents,  since  both  currents  and  voltages  are  similar 
in  form  it  follows  that  the  effective  e.m.f.  shown  by  a  voltmeter 
will  be 


E  = 


Problem  61.  —  In  Fig.  110  let  E  be  the  known  impressed  voltage,  let 
capacity  =  C  farads,  inductance  =  L  henrys  and  resistance  =  r  ohms. 
Then 

Xe  =  ;  XL  =  27T/L. 


Find  the  current,  and  the  voltage  drops  across  the  inductive  impedance 
and  the  capacity,  when  the  impressed  voltage  is  composed  of  a  fundamental 
and  a  third  harmonic.  The  fundamental  component  of  current  will  be 


where 

b  =  - 

Zo" 

and 

Zo2  =  r2  +  (XL—  xc)*, 

and  EI  =  e\  is  the  zero  vector. 


DISTORTED  WAVES.     RESONANCE  EFFECTS    137 


The   voltage   drop   across   the   inductive   impedance,   z,   due   to    /i,  is 

Eiz  =  I\Z  =  ei(g  -{- jb)(r  -\-JXL)  =  a  +  jb' 
where 

a  =  e\gr  —  e\bx]  br  =  (br  +  gx}e\. 

The  voltage  across  the  capacity  reactance  due  to  /i  is 

Eic  =  7i(0  -  jxc)  =  d  +jf, 
where 

d  =  eibxc]  f  =  —  e\gxc. 

The  third  harmonic  components  of  current  and  voltage  are  similarly  deter- 
mined, remembering  that 

x3L  =  SZL, 
_  _?5. 

Problem  62. — In  the  circuit  of  Fig.  110, 
let  r  =  1,  L  =  0.0265,  C  =  0.000265,  E  = 
100,  E3  =  30.  Find  and  plot  the  current 
waves  /i,  73  and  7,  as  the  fundamental  pIG  HQ 

frequency  is  varied. 

NOTE.— Solve  for  frequencies  of  15,  20,  25,  35,  50,  55,  60  ,65,  75,  100. 
Solution. — We  have,  first, 


*T 


Then 
where 

73  = 
where 


-  #32  =       lOO2  -  302  =  95. 
95  95r-a;i)  95r 


95(r-.ysi)  _ 
r2  +  zi2         r2 + 


1 


=    XiL    ~ 

30 


30r 


+ 


+jif3, 


=  x3L  - 


1  600  200 

=  2./3L  -  ^77,  =  0.167/a  -    77    =  0.5/t  -  -jr- 


27T/3C 


Waves  of  7lf  73  and  /  are  plotted  in  Fig.  111. 

It  is  seen  that    the    maximum    /i  occurs  when    XIL 


t  *•«•»   when 


0.167/1  =  -7-.  or,  at  the  frequency  /i  =  60,  and  it  is  Ilm  =  -j  =  95  amp. 

200 
Maximum  73  occurs  when  x3L  =  x3c,  i-e-j  when  0.5/i  =  -7—1  or,  at  the  fre- 

30 
quency  /i  =  20  and  it  is  Ism  =  ~r  =  30  amp. 


138  ELECTRICAL  ENGINEERING 


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DISTORTED  WAVES.     RESONANCE  EFFECTS    139 


40 

Frequency 

FIG.  111. 

Fig.  Ill  gives  a  striking  illustration  of  resonance,  in  that  there  are  two 
distinct  resonance  points. 

Correspondingly  more  points  of  resonance  would  be  produced  if  the 
e.m.f.  wave  possessed  more  harmonics.  The  higher  the  harmonic,  the 
lower  the  frequency  at  which  resonance  occurs. 


CHAPTER  XXI 

CONSTANT  POTENTIAL— CONSTANT -CURRENT  TRANS- 
FORMATION (Cont'd  from  Chapter  XVII) 

It  has  already  been  shown  how  a  fairly  constant  current  may 
be  obtained  from  a  constant  potential  source  by  the  use  of  re- 
sistance and  inductive  reactance. 

It  is  recollected  that  either  the  efficiency  or  the  power  factor  is 
poor,  and  that  the  range  of  fairly  constant  current  is  quite  narrow. 
While  a  far  better  control  can  be  obtained  by  the  introduc- 
tion of  a  condenser  in  connection  with  the  reactance,  this  latter 
method  has  found  little  practical  application  because  of  the 
rather  high  cost  of  condensers  and  their  unsatisfactory  operation 
with  the  distorted  current  taken  by  arc  lamps.  It  is,  however, 
probable  that  such  system  will  be  extensively  used  in  the  future 
on  account  of  the  fact  that  series  incandescent  street  lighting  is 
being  used  to  an  increasing  extent. 

In  Fig.  112,  let  E0  be  the  constant  voltage  impressed  on  the 
system,  and  let  E  be  the  variable  voltage  across  that  part  of  the 
system  in  which  the  constant  current,  7, 
is  to  be  maintained.  Let  the  system  be 
composed  of  z\  =  r\  +  jx\,  in  series  with 
the  parallel  circuits,  z2  =  r2  +  jxz  and  the 


*2f  [       f      variable  load  impedance,  z  —  r  +  jx.     E, 
the  voltage  across  the  variable  impedance, 
FlG    112  will  be  taken  as  zero  vector,  =  e,  although 

it   might    seem    more   logical  to  use  EQ 

which  is  known.  (With  EQ  chosen,  the  calculations  are  some- 
what more  involved,  yet,  of  course,  perfectly  possible,  and  the 
student  is  advised  to  apply  both  methods  and  verify  results.) 

/2  =  -  =  e(g2  +  j62),  (55) 

22 

E0  =  e  +  IlZl  =  e(h+fl). 
140 


CONSTANT-CURRENT  TRANSFORMATION 

Numerically, 

#o  = 
and 


141 


- 


which  is  the  value  of  the  zero  vector. 

Substituting  this  value  of  e  into  (54),  (55)  and  (56), 


(58) 


(60) 


These  equations  give  the  currents  in  the  three  branches  of  the 
circuit,  but  there  is  nothing  about  them  which  determines  that 
7  shall  be  constant.  EQ,  g%  and  62  are  constants,  g  and  b  are 
variable,  and  g0  and  bQ  are,  respectively,  g  +  $2  and  6  +  62. 
h  and  I  are  made  up  of  combinations  of  g,  #2,  b,  62  and  rj.  and  x\, 
where  rx  and  #1  are  constants.  I\  and  72  are  essentially  variable 
because  of  the  variation  of  z,  the  load  impedance. 

Problem  63. — Let  a  circuit  be  chosen  as  in  Fig.  113, 

such  that   7*1  =  r-z  =  0  

Xl  =  -  x,  =  125  ohms 
x  =  0. 
Let 

Eo  =  250  volts. 

Find  e,  /i,  7  and  the  power  factor,  and  plot 
them  against  r  for  varying  values  of  r  from  0 
to  500  ohms.  FIG.  113. 


xz 


Solution. — 
Since 

.'.  (54)  becomes 
Similarly,  (55)  becomes 


x  =  0,  9  =  -'t  b  =  0. 


and  (56)  becomes 


e  I  -  —  7  — 


142  ELECTRICAL  ENGINEERING 

and 

E0  =  250  =  e  + 


whence 


+ -«<*+*>, 


Substituting  values  for  xi  and  z2, 


125  •*"  1252 


(57)  then  becomes 


(58)  becomes 


(60)  becomes 


250 
«-lM-2r- 

r 

/  =  —  =  2  =  constant  for  all  values  of  r. 


+  122 


62.5 


This  case  is,  of  course,  ideal  in  that  it  assumes  absence  of  re- 
sistance in  both  reactive  branches  of  the  circuit. 

In  Chap.  XVII,  it  was  found  that  with  the  system  which  used 
only  inductive  reactance  to  obtain  constant  current,  the  power 
factor  was  quite  low. 

To  obtain  an  expression  for  the  power  factor  in  the  present  case, 

h  jl)  =  eh  +  jel, 


,e 


whence,  by  telescoping,  the  power  is 


_o  n  ^  m 

e  i        x\\       e  x\       e 

=  —  ( 1  -f-      )  —  -       =- 

r  \         xj       x2r        r 


Substituting  numerical  values,  P0  =  4r. 


CONSTANT-CURRENT  TRANSFORMATION 
Volt-amp.  =  #0/1  =  (eh  +  jel)  (-  -  j  -} 

\T  3/2/ 

e%      eH 


143 


Substituting  values  for  h  and  I, 


This  equation  reduces  to 


when  Xi  =  —x^  and,  substituting  numerical  values, 

1  =  4Vr2  +  I252' 


.'.  Power  factor  =W^F~  = 


Tabulating : 


2  +  1252 


T 

o 

10 

20 

50 

100 

200 

500 

e  

0 
0 

20 
100 

40 
400 

100 
2,500 

200 
10,000 

400 
40,000 

1,000 
250,000 

r2  +  1252.  .  . 

15,625 

15,725 

16,025 

18,125 

25,625 

55,625 

265,625 

Vr2  +  1252 
P.F  
/i  
72  

125 
0 
2 
0 

125.2 
0.080 
2.00 
0  16 

126.5 
0.158 
2.02 
0  32 

134.5 
0.372 
2.15 

0  8 

160 
0.625 
2.56 
1.6 

236 
0.847 
3.78 
3.2 

515 
0.97 
8.25 
8.0 

Fig.  114  shows  the  curves  of  current,  voltage,  and  power  factor 
for  change  of  resistance  of  the  load. 

In  this  case,  the  power  factor  is  seen  to  be  very  much  better 
than  it  was  found  to  be  where  inductive  reactance  alone  was  used. 

Problem  64. — Let  the  load  in  the  preceding  problem  be  made  up  of  both 
r  and  x.  Find  the  effect  of  reactance  in  the  load  and  make  a  general  study 
of  the  conditions  under  varying  power  factor  of  the  load. 


144 


ELECTRICAL  ENGINEERING 


This  may  be  done  as  follows:     Imagine  the  load  to  consist  of  any  number 
of  lamps,  each  lamp  possessing  a  certain  resistance  and  a  certain  reactance. 

Then  the  ratio  of  -  will  be  constant. 

1.  Let  x  =  0.5r.     The  power  factor  of  the  load  will  then  be 


1 


•V/r2+0.25r2       Vl.25 


0.895 


2.  Let  x  =  r.     The  load  power  factor  is  then  0.707. 

Supplying  these  values  in  turn  to  Eqs.  54,  etc.,  as  in  the  previous  problem, 
tabulations  and  curves  may  be  obtained  from  which  a  report  on  the  effect 
of  power-factor  variation  of  the  load  may  be  made. 

In  order  to  bring  the  subject  to  a  practical  basis,  the  effects  of 
actual  constants  of  the  apparatus  should  be  investigated.  It 


400  500 


Resistance 
FlG.    114. 


will  be  found  that  the  resistance  of  a  suitable  reactive  coil  for 
such  a  case  as  developed  above,  would  not  need  to  be  above 
0.5  ohm,  and  that  the  resistance  in  the  condenser  circuit  would 
be  very  much  less.  Consequently  the  effects  of  resistance  are 
extremely  small,  and  the  case,  as  worked  out,  may  be  considered 
as  approximately  attainable  in  practice. 

Many  schemes  have  been  proposed  for  the  attainment  of  con- 
stant current  from  a  constant  potential  source,  in  which  more  or 
less  elaborate  combinations  of  reactances  have  been  arranged.  A 


CONSTANT-CURRENT  TRANSFORMATION       145 

study  of  the  possibilities  of  different  schemes  is  profitable  for  the 
student  as  it  affords  excellent  practice  in  circuit  calculation.1 

Power  and  Wattless  Components  of  Volt-amperes. — The 
quantity  El  cos  a  is  called  the  power  component  P  of  the  volt- 
amperes. 

By  a  similar  conception,  El  sin  a  is  called  the  "  wattless" 
component  P1  of  the  volt-amperes. 

Thus 

El  =  V(EI  cos  a)2  +  (El  sin)2- 

Referring  to  Eq.  (36) 
El  sin  a  =  El  sin  (0  -  7) 

=  El  (sin  )8  cos  7  —  cos  0  sin  7) 

=  EI  ^]^  =  i'e  -  e'i.  (61) 

1  For  some  of  these  developments  see  STEINMETZ,  "Alternating-current 
Phenomena,"  Chap.  X. 


10 


CHAPTER  XXII 
THEORY  AND  USE  OF  THE  WATTMETER 

The  most  accurate  way  of  obtaining  results  in  the  measure- 
ment of  alternating  current,  voltage  or  power  is  by  the  use  of  the 
electro-dynamometer. 

As  generally  employed  the  electro-dynamometer,  invented  by 
SIEMENS  is  a  combination  of  2  coils,  one  movable  and  the  other 
fixed,  whose  planes  are  set  at  right  angles  to  each  other.  When 
current  is  sent  through  the  coils,  each  sets  up  a  magnetic  field 
in  the  region  occupied  by  the  other,  thus  causing  forces  which 
tend  to  move  the  coils  relatively  to  each  other.  The  forces  are 
balanced  by  tension  on  a  calibrated  spring. 

If  the  same  current  is  sent  through  both  coils,  the  scale,  prop- 
erly calibrated,  measures  the  current.  If  the  instrument  is 
placed  in  series  in  a  circuit,  the  current  measured  is  that  of  the 
circuit,  and  the  meter  becomes  an  ammeter.  If  it  is  placed  in 
shunt  to  a  given  circuit,  the  current  in  the  coils  is  proportional 
to  the  voltage  drop  in  the  circuit,  and  the  meter  becomes  a  volt- 
meter. 

If,  however,  one  coil  is  placed  in  series  and  the  other  in  shunt 
to  a  given  circuit,  the  effect  on  the  instrument  is  proportional  to 
the  product  of  the  amperes  and  the  volts  at  any  instant,  and  the 
electro-dynamometer  becomes  a  wattmeter  and  measures  power. 1 
For  practical  construction,  the  coil  to  be  connected  in  series  is 
made  of  few  turns  of  comparatively  heavy  wire,  and  is  usually 
the  fixed  coil,  while  the  coil  to  be  connected  in  shunt  is  made  of 
very  many  turns  of  fine  wire  and  is  movable. 

Accurate  results  are  obtained  by  the  dynamometer  because  the 
coils,  while  readings  are  made,  are  always  kept  in  the  fixed  rela- 

1  The  flux  set  up  by  1  coil  (fixed)  is  proportional  to  the  current  flowing 
in  the  circuit,  while  the  flux  set  up  by  the  other  coil  (movable)  is  proportional 
to  the  voltage  across  the  circuit.  But  the  force  at  any  instant  acting  on  the 
coils  is  proportional  to  the  product  of  the  fluxes  set  up  by  the  coils,  that  is, 
the  force  on  the  coils  is  proportional  to  the  product,  E  X  /,  where 

E  =  voltage  across  the  circuit,  and 
/  =  current  in  the  circuit. 
146 


THEORY  AND  USE  OF  THE  WATTMETER       147 


tive  position  at  right  angles  to  each  other,  thus  eliminating  mutual 
flux;1  also  because  no  iron  is  used  in  construction,  and  there  are 
no  other  materials  which  might  cause  variation  in  the  results. 

Accuracy  must  be  obtained,  however,  by  the  correction  of 
certain  errors  in  the  readings.     The  error  due  to  friction  of  the 
movable  coil  is  small.     The  error  due  to  changes  of  resistance 
by  temperature  is  obviated  in  good 
instruments   by  the  use   of  resis- 
tance  which    is    not    affected    by 
change    of    temperature.      When 
used  as  a  wattmeter,  the  readings 
of  the  dynamometer  must  be  cor- 
rected for  error  due  to  the  manner 
of  connection  in  the  circuit.     This 
correction  is  of  great  importance. 

Let  the  wattmeter  be  connected 
as  in  Fig.  115  (A),  in  which  the 
power  consumed  by  the  impedance, 
Z  =  R  +  jX,  is  to  be  measured. 

The  current  coil  is  represented 
by  the  impedance  z  —  r  +  jx]  the 
voltage  coil  by  the  impedance  z\  =     < 
7*1  -\-jx\.     The  load   voltage  is  e, 
which  is  chosen  as  the  zero  vector. 


(B) 


(c) 

FIG.  115. 


The  meter  should  be  first  calibrated  by  direct  current,  so  that 
its  reading  for  any  given  direct-current  power  is  known. 

Wattmeter  Connections. — Connection  (A)  is  wrong,  because 
the  current  coil  has  to  carry  /i,  the  current  taken  by  the  voltage 
coil,  in  addition  to  the  load  current  /,  thus  causing  the  wattmeter 
to  indicate  the  power  lost  in  the  voltage  coil,  plus  the  load.  There 
will  thus  be  a  reading  even  at  no-load.  If  JiVi,  the  power  lost  in 
the  voltage  coil,  is  subtracted  from  the  wattmeter  reading  this 
error  is  eliminated.  This  error  is  usually  negligible  in  connection 
with  circuits  carrying  large  current  at  low  voltage. 

In  the  connection  shown  in  Fig.  115  (B),  the  current  coil 
carries  only  the  load  current.  The  voltage  coil,  however,  is  so 
connected  as  to  include  the  drop  in  the  current  coil  as  well  as 
that  across  the  load.  Thus  the  wattmeter  indicates  the  power 


1  Mutual  induction  will  be  treated  more  fully  in  connection  with  the 
study  of  the  transformer.     See  Chap.  XXVI. 


148  ELECTRICAL  ENGINEERING 

lost  in  the  current  coil  in  addition  to  the  load.  This  error  may  be 
corrected  by  subtracting  the  Io2r  power  lost  in  the  current  coil. 
Connection,  (B),  is  best  adapted  to  measurements  at  high  voltage 
and  low  current. 

A  third  arrangement,  Fig.  115,  (C),  is  known  as  the  compen- 
sated wattmeter.  In  this  there  is  wound  a  fine  wire  coil  of  the 
same  number  of  turns  as  the  current  coil  directly  upon  the  latter. 
By  its  connection,  it  is  seen  that  this  coil,  c,  carries  the  current  of 
the  voltmeter  coil.  It  therefore  supplies  to  the  current  coil  just 
enough  back  ampere-turns  to  neutralize  those  due  to  that  excess 
of  current  in  the  current  coil.  This  arrangement  causes  the 
wattmeter  to  read  correctly  so  far  as  its  connections  are  con- 
cerned. 

Readings  of  the  dynamometer  calibrated  by  direct  current 
must  also  be  corrected  for  an  error  due  to  change  in  frequency. 
In  connection  (A),  Fig.  115,  the  load  current  is 


Current  in  the  voltage  coil  is 

e 

Current  in  the  current  coil  is 

/o  =  7  +  1 1  =  e(gQ  +  jb0). 

\        These    currents    are   plotted   in   Fig. 

"el!6.  Usually,  the  angle  of  lag  of  /i,  is 
very  small,  even  less  than  1°.  This 
slight  lag  may,  however,  give  a  large 
error. 

Tan  7  =  —  • 

Since  the  coils  are  at  right  angles,  the 
torque  at  any  instant  on  the  movable  coil  is  proportional  to 
the  product  of  the  currents. 
Thus, 

T  =  ki0i1} 
where  iQ  and  ii  are  instantaneous  values  of  current  in  the  two  coils. 


THEORY  AND  USE  OF  THE  WATTMETER       149 

Let 

io  =  I0m  sin  coJ,  and  ii  =  Ilm  sin  (ut  +  a). 

Then  the  average  value  of  the  torque  through  one-half  cycle  is 

fc/« 

-  I  (/Ow  sin  ut)(Iim  sin  (co<  -f  a)dt 


/i  A  . 
I  (si 


. 
sin2  co<  cos  a  +  sin  ut  cos  w£  sin  a)dt 


o/]  Ah  -  cos2o;£  sin2co£  1 

—  I  ~~2~~    ~  COS  a  "^  ---  2  —  Sln  a 

2fc/o/i["fa)<  cosa  __  cos  a  sin  2co^       sin  a  cos  2<on* 
TT     L      2  4  ~4  Jo 

7r  sin  a    ,    sin  af|       2fc  _ 

cos  a  --  -  +  --    =        70/i  cos  a 


cos  a, 

/o  and  7i  being,  as  usual,  the  effective  values. 
The  wattmeter  reading  is  then  proportional  to 

/i/o  cos  a, 

where  a  is  the  angle  between  J0  and  /i. 

But  the  true  power  is  el  cos  ft  where  ft  is  the  angle  between 
e  and  /. 

In  order  that  the  wattmeter  shall  read  true  power  it  is  therefore 
necessary  to  correct  each  term  in  the  reading.  These  corrections 
are:  (a)  for  the  current  in  the  voltage  coil;  (6)  for  the  currerit  in 
the  current  coil;  (c)  for  the  angular  displacement. 

(a)  Assuming  the  meter  to  be  calibrated  by  direct  current,  the 
current  in  the  voltage  coil  at  any  frequency,  /,  will  be  less  in  the 

ratio     /-—--=>   due  to  the  inductance  of  the  coil. 

V  ri2  +  Zi2 

The  reading  should  therefore  be  corrected  by  the  factor  —  » 
in  order  to  bring  it  proportional  to  e. 

(b)  The  current  in  the  current  coil  is  too  large  in  the  ratio,  -j  • 
The  correction  factor  is  therefore, 

L 

/o 


150  ELECTRICAL  ENGINEERING 

(c)  The  correction  factor  for  angular  displacement  is  evidently 

^-^i  where  0  and  a  are  as  shown  in  Fig.  116.     The  complete 

cos  a 

correction  factor  is  then 

21    IG2  +  B2  cos  ft 

The  constants  of  the  wattmeter  are  assumed  known,  which  per- 
mits of  obtaining  all  angles  except  the  phase  angle  of  the  load. 
Thus,  a  is  known,  but  not  0.  In  order  to  obtain  0,  a  reading  may 
be  taken  of  the  wattmeter,  voltmeter  and  ammeter  as  in  the  ex- 
ample below.  Then,  roughly, 

=  ~-  (62) 


Substituting  this  value  of  cos  0  into  the  correction  factor,  a 
new  value  of  W  is  obtained.  Replacing  the  approximate  W  of 
(62)  by  this  new  value,  a  new  value  of  cos  0  is  obtained  in  which 
the  error  is  of  the  second  magnitude.  By  repeating  this  process 
any  desired  degree  of  precision  may  be  obtained. 

Example. — To  find  cos  /3,   when  by 
reading    of    instruments   the   approxi- 


mate power  factor  is  found  to  be 


=  _  =    50 

volt-amp.         100 


pow^r  Factor'25  There   must   be   a   correction-factor 

Fm.  117.  curve  of  the  dynamometer  for  varying 

power  factor. 

From  this  curve,  let  the  value  of  k  be  0.99  f or  P.F.  =  0.5.  Then 
multiplying,  0.5  X  0.99  =  0.495  =  power  factor  to  second  ap- 
proximation. It  is  evident  that  a  repetition  of  the  process  will 
be  hardly  necessary  in  most  practical  cases. 

Problem  66. — With  the  wattmeter  connected  as  above  (115,  A),  determine 
and  discuss  the  correction  factors:  (1)  with  non-inductive  load;  (2)  when  the 
power  factor  of  the  load  is  just  equal  to  the  power  factor  of  the  voltage 
coil;  (3)  in  the  theoretical  case  when  there  is  no  self-induction  in  the  voltage 
coil. 

Problem  66. — By  a  process  similar  to  that  just  given,  find  the  correction 
factors  for  wattmeters  when  connected  according  to  (Fig.  115,  B  and  C). 

The  errors  actually  obtaining  in  practice  with  good  commercial 
indicating  wattmeters  are  quite  small.  Thus,  at  normal  voltage, 


THEORY  AND  USE  OF  THE  WATTMETER       151 

2000  cycles  and  power  factor  from  0.8  leading  to  0.8  lagging, 
the  error  is  usually  less  than  J4  per  cent. 

As  the  power  factor  is  lowered  the  error  becomes  larger.  At 
normal  voltage,  60  cycles,  the  error  may  be  less  than  0.2  per  cent, 
with  the  power  factor  down  to  0.1. 

If  the  impressed  voltage  is  low,  say  15  per  cent,  of  normal,  the 
error  may,  however,  be  several  per  cent. 

In  operation  there  are  also  errors  which  enter 

with  the  use  of  "current"  and  "potential"  trans-        j|  g 

formers.     When   these  are  used,   the  error  is  |^iovoit    5 

practically  negligible  for  power  factors  above  f   '        _j 

0.8  except  for  small  loads.  With  non-inductive- 
load  of,  say  10  per  cent,  normal,  the  error  may, 
however,  be  several  per  cent. 

Problem67. — An  uncompensated  wattmeter  (Fig.  115,  A  &ndB  and  Fig. 
118)  has  a  rating  of  400  watts.  At  100  volts  the  resistance  of  its  voltage 
coil  is  2000  ohms.  At  50  volts  the  resistance  is  1000  ohms,  at  10  volts  it  is 
200  ohms.  The  inductance  of  the  voltage  coil  is  0.007  henry.  Resistance 
of  the  current  coil  is  0.03  ohm.  Inductance  of  the  current  coil  is  0.0003 
henry.  Find  the  wattmeter  reading,  the  actual  watts  and  the  correction 
factor  for  all  combinations  of  voltage,  current  and  power  factor,  when 

e  =  100,  50.0,  and  10.0,  volts 

7  =       4,  and     0.4  amp., 

P.F.  =       1,  0.1  lead,  and    0.1  lag. 

/  =60  cycles. 


CHAPTER  XXIII 
SIMPLE  PROBLEMS  IN  ELECTRO -STATICS 

It  is  desirable  at  this  point  to  introduce  certain  principles  of 
electro-statics.  These  should,  of  course,  be  more  or  less  familiar 
to  every  student  who  has  had  an  adequate  course  in  physics. 

Potential. — By  definition,  the  potential  at  a  point  in  an  electric 
field  is  equal  to  the  work  done  per  unit  charge  in  bringing  a 
positive  charge  from  a  place  of  zero  potential  (usually  infinity) 
to  the  point. 

Intensity. — Also  by  definition,  the  intensity  of  the  electric 
field  (lines  per  square  centimeter  in  air)  is  numerically  the  same  as 
the  force  which  that  field  exerts  on  unit  charge. 


FIG.  119. 


Thus,  if  R  is  the  intensity  of  the  field  at  a  distance  r  from  a 
point  charge,  Q  (Fig.  119).     See  also  Chap.  XVIII. 


R   =  _  _  =  =  Q 

area  of  sphere  of  radius  r       4.irr2       r2 

Therefore  the  potential  at  p  is 

VP  =  -  fRdr  =  -  fR  cos  6ds  (63) 

where  ds  is  an  element  of  the  path  of  the  unit  charge  and 

dr  —  ds  cos  6. 

The  minus  sign  is  used  because  work  is  done  in  bringing  unit 
positive  charge  against  the  charge  Q  which  is  also  assumed  posi- 
tive. Thus,  the  repulsion  between  the  charges  must  be  overcome 

152 


SIMPLE  PROBLEMS  IN  ELECTRO-STATICS      153 

and  work  has  to  be  supplied.     This  designation  is,  of  course,  a 
matter  of  convention. 

Substituting  the  value  of  R,  just  obtained, 


VP  =  -    1    Rdr  =  -    \    Qdr  = 

J-  Jj2  r_       p 


Q 


P      Q 


where  p  is  the  distance  from  Q  to  p. 

Capacity  of  a  Sphere. — Suppose  the  charge,  Q,  to  be  on  an  iso- 
lated sphere  of  radius,  n.  Then,  at  the  surface  of  the  sphere 

p  =  ri,  and  the  potential  is  Vi  =  —• 

The  capacity  of  a  condenser  is  defined  as  the  charge  per  unit 
potential.  Thus,  C  =  y,  where  C  is  capacity,  and  V  is  the 
potential  of  the  charge  Q. 

Since,  therefore,  with  an  isolated  sphere,  Vi  =  —  i  the  capacity 
of  the  sphere  is 


in  cm. 


Thus,  the  capacity  of  a  sphere  is  numerically  equal  to  its  radius; 
the  value  of  the  capacity  expressed  in  farads,  C  is  found  by 
dividing  C  in  centimeters  by  the  constant  9  X  1011. 

Potential  Gradient.  —  The  potential  gradient,  usually  denoted 
by  G,  or  the  rate  at  which  the  potential  changes  at  a  given  point, 
is  of  very  great  practical  importance  since  it  is  a  measure  of  the 
electric  stress  to  which  the  dielectric  is  subjected.  The  potential 
gradient,  G,  and  the  electric  field  intensity,  R,  are  the  same  nu- 
merically. Thus,  if  the  potential  of  a  certain  point  falls  at  the 
rate  of  5  units  of  potential  per  cm.,  the  actual  number  of  lines 
per  sq.  cm.  at  the  point  is  also  5. 

By  definition, 


Since, 

dV  =  -  Rdr 


. 

In  a  dielectric  of  specific  inductive  capacity,  K,  the  intensity 
as  well  as  the  potential  gradient  for  a  given  charge  is  less  than  in 


154  ELECTRICAL  ENGINEERING 

air.     It  is  -  times  the  intensity  in  air.     Thus,  in  the  case  of  a 
sphere, 

c      v      l  Q 

G==  R==  -  -,- 

The  maximum  possible  value  of  G,  or  R,  under  ordinary  con- 
ditions in  air,  is  not  known  exactly,  but  is  in 
the  neighborhood  of  30,000  volts  per  cm.,  or 
100  electro-static  units  of  potential. 

Capacity  of  a  Spherical  Concentric  Con- 
denser.  —  Consider  2  spherical  concentric  bodies 
with  charges  plus  and  minus  Q. 

By  (63),  the  potential  difference  is 


r*  =  r 
=  -  Rdx, 

Jx  =  n 


where  r  and  n  are  ^radii  respectively  of  the  inner  and  outer  sur- 
faces of  the  condenser. 
But 

Q 


--  r 

t/  n 


Therefore  the  capacity  of  the  condenser  is 

C  =  Q       -?^-in  cm.  (65) 

ri  -f-  r 

Potential  gradient  between  concentric  spheres. 
Since 

rfF 

G  = T-> 

dr 

and 

dV  =   -  Rdr, 


But 


SIMPLE  PROBLEMS  IN  ELECTRO-STATICS      155 

rr« 

c  = 


— r 


At  the  surface  of  the  smaller  sphere,  x  =  r,  whence  the  gra- 
dient is 

7*1  V 

=  7  (r,  -  r)' 

The  Capacity  of  a  Concentric  Cylinder.— Let  the  charges  be 
±  Q  per  cm.  of  length  of  the  cylinder  (Fig.  121). 
Then,  by  GAUSS'  theorem,  the  flux  emanating 
from  each  centimeter  of  length  =  4irQ. 

Lines  of  flux  are  here  assumed  to  extend  radi- 
ally, which  they  actually  do. 

At  any  distance,  x}  from  the  center  of  the 
cylinder  the  intensity  at  a  point  is  the  total 
number  of  lines  divided  by  the  area,  or, 


Thus,  the  potential  difference  is: 

dx 
+  2Q  log  £ 


r        c^Q 

—  I    Rdx  =  —    I   —  dx  =  —  2Q[log  r  —  log  rj 
Jn  -Jri 


(66) 


and  the  capacity  is 


n 

C  =  —  = 


in  cm. 


per  cm.  length  of  the  concentric  cylinder. 
The  gradient  at  any  distance,  x,  from  the  center  is 


x          x         x  "  _ .      r,  .     r,         (68) 

2  log  -        x  log  - 


At  the  surface  of  the  inner  conductor,  x  =  r. 

a 


r  log  - 
&  r 


i  and  this  is  the  greatest  value  of  the  gradient. 


1  See  "Advanced  Course  in  Electrical  Engineering." 


156 


ELECTRICAL  ENGINEERING 


In  these  formulae  no  account  is  taken  of  any  effects  due  to  the 
ends  of  the  concentric  cylinder.  For  the  special  case  of  an  outer 
cylinder  of  radius  ri  =  °° ,  C  =  0. 

Capacity  of  Two  Parallel  Plates  so  Large  that  the  Effects  of 
Their  Edges  may  be  Neglected. — The  total  flux  set  up  by  a 
charge,  Q,  is  IwQ  (Fig.  122). 

4x0 
The  intensity,  R  —  —j-> 

where  A    is  the  area  of  one  side  of 
the  plate. 

The  potential  difference  is: 


A 

t    +Q 

1   f 

/ 

^ 

1 

1 

X1 

^    -Q 

1 

V 

JS 
FIG 

122. 

-$> 


*>  '  - 


where  d  is  the  distance  between  the  plates. 
The  capacity 

Q        A         KA 


C  =  ~~  =  ~4 — j  =  -A — ;  in  cm. 
e       4ird       4?ra 


(69) 


(70) 


where  the  dielectric  has  a  specific  capacity  K. 

The  potential  gradient,  Gy  is  a  constant  in  the  dielectric  be- 
tween the  plates,  since  the  flux  lines  are  parallel. 
Thus, 

C  -       —  —  ^Q  _  ^TrCe  _  efc 
~  dx  =  ~~A  A      =  ~A 

in  which  e  is  the  difference  of  potential  of  the  plates  and  k  is  a 
constant,  =  4irC. 

Capacity  of  a  Transmission  Line.1 — The  line  is  represented  in 
section  in  Fig.  123,  with,  r,  the 
radius,  and,  D,  the  distance  be- 
tween centers,  of  the  wires  A  and 
B.  Let  A  be  charged  +  Q,  and"*"0 
B}  —  Q.  The  flux  lines  emanat- 
ing from  A  enter  B.  The  inten- 
sity at  a  point,  p,  due  to  the  charge  on  A,  is  R A;  that  due  to 
the  charge  on  B  is  RB. 


*|^r 


I 


FIG.  123 


1  For    more    exact    deduction    see    "Advanced     Course     in     Electrical 
Engineering." 


SIMPLE  PROBLEMS  IN  ELECTRO-STATICS      157 
Then 


r>  --"  v&        ^vl 

RA  -  ^z  =  — 


p 

" 


2ir(D-x)  ~~  (D-x) 


The  intensity  due  to  the  two  charges  is  the  sum  of  RA  and  RB, 
since  the  direction  of  the  lines  of  electro-static  force  from  A  ,  due 
to  a  positive  charge,  is  the  same  as  that  due  to  B,  which  has  a 
negative  charge. 


The  potential  difference  is: 

Cr  CT     /I  1      \dx 

e  =  -  \  Rdx  =  -  2Q  I        (-  +  ~  -) 
JD-T  JD-T  \*       D  ~  x/ 

=  4Q  log  ^—^  (71) 

and  the  capacity  is  therefore 

1 

(72) 


per  cm.  length  of  circuit  not  of  wire. 

This  capacity  is  expressed  in  centimeters.  If  the  line  is  in  a 
dielectric  of  specific  inductive  capacity,  AC,  the  capacity  in  air  as 
determined  above,  must  be  multiplied  by  K. 

To  transform  capacity,  expressed  in  electro-static  units  in  (72), 
into  electromagnetic  units,  the  former  should  be  multiplied  by 

—2  where  v  is  the  velocity  of  light  =  3  X  10  10  cm.  per  sec. 
The  practical  electromagnetic  unit  of  capacity  is  the  farad. 

C 
Capacity  in  farads  =  -^  X  109, 

where  C  is  capacity  expressed  in  electro-static  units. 
.'.  Farads  =  electro-static    units  X 


..Q//      9  \/ 
Thus,  C/cm.  of  circuit,  in  farads,  = 
k 


4  log  ^~  X  9  X  10"       (4  logio  ^f-r)  X  9  X  10" 


158  ELECTRICAL  ENGINEERING 

When  connected  to  a  source  of  alternating  e.m.f.,  the  effective 
value  of  the  charging  current  is  Ic  =  2ir{CE,  where  E  is  the  effect- 
ive value  of  the  line  voltage. 

The  voltage  is  frequently  taken  from  one 

~E~n  "  side   of  the  line  to  neutral,  that  is,  to  the 

E  ---  *  --------   point  of  zero  potential  of  the  system  (Fig. 

|  _    124).     When  this  voltage  to  neutral  is  used, 

FIG.  124.  the    capacity   to   ground,    or   to   neutral,    is 

twice  as  great  as  the  capacity  between  lines. 

This  follows  since  Ic  =  2irfCnEnj  where  Cn  and  En  are  capacity 

E 

and  voltage  to  neutral,  and  for  single  phase  systems,  En  =  ^r- 

z 
• 

For  three-phase  systems,  E 
k 


2  X  9  X  1011  X  log 


_  ,  farads  per  cm.  of  line,  since  in 


r 
using  the  neutral,  the  length  of  line  is  the  transmission  distance. 

0  0074 
Reducing  values  to  practical  units,   Cn/1000'  =  -        ^  _ 

logic  — — 

is  the  capacity  to  neutral  per  1000  ft.  of  line,  in  micro-farads. 
7C/1000'  =  — 1Qn6      is  the  charging  current  per  1000  ft.  of  line,  in 

amperes. 

Capacity  of  a  Three-phase  Cable. — Capacity  to  neutral  per 
1000  ft.  of  line  is  given  in  micro-farads  by  the  formula 

0.0074  i 


Cn/1000' 


V3a  R*  -  a2 

logio 


r  R*  +  a4  + 


Such  a  cable  is  represented  in  section  in 
Fig.  125,  where  R  is  the  radius  of  the  sur- 
rounding sheath,  a  is  the  distance  from  the 
center,  or  neutral  point,  to  the  center  of  one  FlG-  125> 

of  the  wires  and  r  is  the  radius  of  1  wire. 

Problem  68. — (a)  Prove  that  the  greatest  charge  which  may  be  put  on  a 
ball  of  10  cm.  radius  is  10,000  electro-static  units.     (Assume  that  the  maxi- 

1  This  will  be  understood  from  later  discussion  of  polyphase  systems  and 
deduced  in  the  volume  dealing  with  advanced  electrical  engineering. 


SIMPLE  PROBLEMS  IN  ELECTRO-STATICS      159 


mum  gradient  is  30,000  volts  per  cm.  when  air  at  atmospheric  pressure 
"breaks  down"  and  a  glow  called  corona  appears  around  the  wire.) 

(6)  Prove  that  the  greatest  surface  charge,  in  coulombs  per  sq.  cm.,  is  ~ffp' 

(c)  Show  that  if  the  inside  conductor  of  a  concentric  cable  has  a  radius 
of  1  cm.,  and  the  outside  conductor  is  2  cm.  in  radius,  0.0027  coulombs 
must  be  put  into  1  mile  of  cable  to  cause  it  to  glow  (corona).  Show  that 
the  potential  difference  between  the  2  conductors  is  20,800  volts. 

Inductance  of  a  Concentric  Cable. — The  inductance  is  recol- 
lected to  be  the  interlinkages  of  the  flux  and  turns  per  unit  current. 

In  general,  if  the  m.m.f .  acting  in  a  circuit  is  F  then  the  flux 

4iTrF  X  area  of  magnetic  circuit 
length  of  magnetic  circuit 

The  interlinkage  factor  is  the  fraction  of  the  total  current  en- 
closed by  the  flux,  and 


is 


L  =  7  S  flux  X  interlinkage  factor. 


(73) 


Consider  first  the  flux  in  the  inside  conductor  due  to  the  as- 
sumed uniform   distribution  of   the   current 
in  it. 

At  a  distance  x  from  the  center  (Fig.  126), 

TTX2 

the  m.m.f.  is  — ^  /  where  I  is  the  total  current. 

?rr2 

The  area  enclosing  the  flux  per  centimeter 
length  of  conductor  is  dx  and  the  length  of 
the  magnetic  circuit  is  2irx 

x*  ,dx_    _         x_ 
• .  Wi  =  47T  r,  I  2wx  -     L  r,  a 

TTX2 

This  flux  interlinks  with  — ;  of   the   total    current;   thus   the 


FIG.  126. 


?rr 


u 

interlinkage  factor  is  — 5- 


FIG.  127. 


(Assuming  that  /*  =  1)      (74) 

dXl  Between  the  conductors,  the  flux  inter- 
links with  the  whole  current  (Fig.  127). 
Thus  by  a  similar  reasoning  we  get : 

•fr  =  21'*f 


160  ELECTRICAL  ENGINEERING 

The  current  in  the  inner  conductor  interlinks  with  the  entire 
flux  which  is  in  the  outer  conductor  but  which  is  caused  by  the 
difference  in  m.m.f.  in  the  inner  and  outer  conductors. 

At  distance  x0  the  m.m.f.  is  thus 


__ 
" 


RQ2  -  R2  R,2  -  R2 

The  interlinkage  of  this  flux  with  the  current  in  the  inner  con- 
ductor is,  of  course,  unity,  thus 


The  inductance  of  the  outer  conductor  should  be  added  to  give 
the  total  inductance  of  the  cable. 
The  m.m.f.  is  shown  above  to  be 

j.  R<>2  -  x<>2 


R,2  -  R2 

(Ro2  -  x,2) 
(Ro2-  R2)2 


2 
' 


1  R02  +  R2          2Ro2R2  Ro 

2  RQ2  -  R2  +  (#o2  -  ^R2)2     g  R 

The  total  inductance  L  =  LI  +  L2  +  L3  +  L4  which  is  readily 
proven  to  be 

1       01      R  ,         2/V  Ro       1  3^o2  -  R* 

L  =  -2  +  21og7  +  ^g—^kg  -  -  -   R^  _  R2  cm. 

This  inductance  is  expressed  in  the  absolute  system  of  units. 
By  dividing  by  109  the  inductance  is  expressed  in  henrys. 

Problem  69.  —  Prove  that  there  is  no  flux  outside  of  the  sheath,  the  flux 
set  up  there  by  the  current  in  the  sheath  being  exactly  neutralized  by  the 
flux  set  up  in  the  same  space  by  the  oppositely  directed  current  in  the 
inner  conductor. 

Inductance  of  a  Transmisson  Line.  —  Let  a  transmission  line 
be  represented  as  in  Fig.  128  by  2  conductors,  A  and  B,  of 
radius  r.  Let  the  distance  between  their  centers  be  D.  Each 
conductor  surrounds  itself  with  flux  lines,  the  directions  of  which 
are  indicated  by  arrows.  The  flux  through  any  zone  of  width, 
dx,  between  the  conductors,  due  to  the  current  in  A,  is 


SIMPLE  PROBLEMS  IN  ELECTRO-STATICS      161 

where  x  is  the  distance  of  the  zone  from  the  center  of  A,  and 
Fx  is  the  m.m.f.  due  to  A. 

Similarly,  the  flux  through  dx,  due  to  the  current  in  B  is 


The  flux  due  to  both  A  and  B  is  then 


/  / 

The  inductance  due   to  the   /  I  / 
interlinkages  of  the  conductors  f   I  j 
with  the  flux  between  them  is  \  1  \ 
then,  since  Fx  =  I  in  this  case,     \\ 

V 


,      D-r  4M  .      D  -  r  , 

=  4/i  log  —  ^—  '  cm.  or  j^  log  —  -  —  henries  per  cm. 

To  determine  the  total  inductance  per  centimeter  length  of 
circuit,  that  due  to  mterlinkage  within  the  material  of  each  con- 

ductor must  be  added.     This  has  been  found  (74)  to  be  ^  for 

2 
each  conductor.     Therefore,  the  total  inductance  is 

L  (total)  =  4;u  log  —  --  +  n  cm.  per  cm.  of  circuit     (75) 

In  practical  formulae,  this  becomes 

L  =  0.000015  +  0.00014  logio^- 


in  henrys  per  1000  ft.  of  wire,  not  1000  ft.  of  circuit. 

Note  that  if  the  capacity  between  transmission  lines  is  given 


in  farads  and  the  inductance  in  henrys  ~7ffi  ^  on^  verv 

less  than  the  velocity  of  light  which  is  3  X  1010  cm.  per  sec.  or 
187,000  miles  per  second. 

Problem  70.  —  Explain  the  effect  of  increasing  the  size  of  the  wire  on  the 
inductance  of  a  transmission  line. 

Similarly,  explain  the  effect  of  increasing  the  distance  between  the  wires. 


11 


CHAPTER  XXIV 
DISTRIBUTED  INDUCTANCE  AND  CAPACITY 

In  the  electric  and  magnetic  problems  dealt  with  so  far  it  has 
been  assumed  that  the  electro-static  and  magnetic  fields  propa- 
gate with  infinite  velocity.  In  other  words,  it  has  been  assumed 
that  the  instantaneous  values  of  the  currents  and  e.m.fs.  are  the 
same  at  all  points  of  the  circuit.  This  of  course  is  practically 
true  except  in  very  long  transmission  lines,  since  the  propagation 
of  the  electric  and  magnetic  fields  in  a  dielectric  such  as  air  is  the 
same  as  that  of  light,  or  very  nearly  3  X  1010  cm.  per  sec.  or 
187,000  miles  per  sec.,  and  along  a  transmission  line  it  is  re- 
tarded only  a  small  percentage  due  to  the  fact  that  the  current 
is  not  confined  to  the  surface  of  the  conductor. 

Assuming,  however,  that  the  transmission  line  is  very  long, 
say  300  miles,  then  the  time  interval  between,  say,  the  maximum 
,  ,  value  of  the  current  at  the  beginning 

dl:  "1  r~^*  and  the  end  of  the  line  is  evidently 
;Hj20  sec->  corresponding  in  a  60-cycle 
-« —  system  to  approximately  one-tenth  of 
one  cycle,  or,  approximately,  36°  in 
FIG'  129'  time  phase. 

It  is  thus  seen  that  in  a  long  transmission  line  not  only  do  the 
instantaneous  values  of  the  currents  and  e.m.fs.  vary  from  instant 
to  instant,  but  at  a  given  instant  the  values  of  the  currents  and 
e.m.fs.  are  different  at  different  points  of  the  line. 

This  problem  has  been  treated  very  completely  by  many 
authorities.  The  simplest  solution  appears  to  be  that  by  STEIN- 
METZ,1  which  is  largely  followed  in  the  succeeding  paragraphs. 

Let  Fig.  129  represent  a  long  transmission  line.  Let  r0  =  re- 
sistance per  unit  length  of  line,  XQ  =  reactance  per  unit  length 
of  line,  </o  =  leakage  conductance  per  unit  length  of  line,  60  = 
capacity  susceptance  per  unit  length  of  line,  r  =  rj,  =  total 
resistance  of  the  line.  Let  dl  be  any  small  section  of  the  line. 

Then  assuming  sine  wave  of  current  when  complex  representa- 

1  "Electric  Discharges,  Waves  and  Impulses." 

162 


DISTRIBUTED  INDUCTANCE  AND  CAPACITY  163 


tion  can  be  used,  the  current  entering  this  section  is  /  +  dl. 
The  current  leaving  the  section  is  /. 

Then  /  +  dl  -  I  =  E(gQ  +  jbo)dl  is  the  small  difference  in 
current  in  passing  through  dl,  or  is  the  combined  leakage  and 
capacity  current  across  the  section  of  width  dl.  This  may  be 
written 

dl  =  EY<41  (76) 

Likewise,  dE  =  7(r0  +  jxQ)dl  is  the  e.m.f.  consumed  by  the  re- 
sistance and  reactance  of  the  section  dl,  or 

dE  =  IZodl  (77) 

(76)  and  (77)  become,  then, 

dl 


_ 


Differentiating  these, 


dE 
~dl 

d2E 


(78) 


dl 


dl 


dE 


Substituting  values  of  -rr  and  —r  from  (78), 


dE 
—jr 


(79) 


Thus,  the  second  differentials  of  E  and  /  are  found  to  be  pro- 
portional to  E  and  /,  respectively.  Since  the  two  equations  are 
similar,  their  solutions  are  similar,  differing  only  in  integration 
constants. 

The  equations  (79)  are  of  the  form 


whose  solution  is 


y  =  Ae 


164  ELECTRICAL  ENGINEERING 

The  equation  of  the  current  then  becomes 

or  if,  for  the  sake  of  briefness,  YoZQ  =  v 
I  =  Ach  +  Be'1' 
We  also  have,  from  (78), 


Ti  1 


Differentiating  (81), 


~  =  Ave*  - 
dl 


Substituting  this  into  (82) 


E  =  ~  [Avelv  - 

-*  0 


(81) 
(82) 

(83) 


(84) 


Representing  the  exponentials  of  (81)  and  (84)  in  series, 


Substituting  these  into  (81)  and  (84)  gives: 

727,2  72,,2 

7  =  A  +  Alv  +  A  --  +  .    .    .  +  B  -  Blv  +  B  ~-  - 


=  A  +  B  +  lv(A  -B)  +  --(A  + 

i        nv\ 
=  (A  +  B)  (l  +  ^)  +  (A  -  B)lv 

and,  similarly, 


(85) 


If  I  is  made  any  length  counting  from  the  receiving  end  of  the 
line  at  I  =  0,  E  =  e,  the  receiving  end  voltage,  and  /  =  /i,  the 
load  current,  both  of  which  may  be  known. 


DISTRIBUTED  INDUCTANCE  AND  CAPACITY  165 

Substituting  these  values  (85)  becomes,  for  the  receiving  end, 
/i  =  A  +  B 


By  substituting  these  values  of  II  and  e  in  (85)  we  get  finally: 


(87) 


where  /  and  E  are  the  values  of  current  and  voltage  at  any  point,  /, 
along  the  line.  The  current  and  voltage  at  the  generator  are 
found  by  substituting  in  (87) 

7  .,  7V 

1/2^2          &  1 

~2~      ^T 

where  Z  and  Y  are  the  values  of  impedance  and  admittance  for 
the  entire  line.  The  equations  (87)  become 


,         ™  (88) 

#o  =  e(. 

YZ  is  found  from  the  constants  of  the  line,  thus: 

YZ  =  (g  +  jb)  (r  +  jx)  =  gr  +  gjx  +  jbr  -  bx 
=  gr  —  bx  +  j(gx  +  br). 

Problem  71. — Transmission  Line  Calculation. — A  200-mile,  three-phase 
transmission  line  is  composed  of  three  No.  000  B.  &  S.  wires,  and  runs  at  an 
altitude  of  1200  ft.  where  it  may  be  assumed  that  the  corona  loss  is  1  kw.  per 
wire  per  mile  at  a  potential  difference  of  125,000  volts  between  the  lines. 

Let  E  =  125,000  volts  at  receiving  end,  between  wires; 
/  =  60  cycles;  r  =  64  ohms  per  wire; 
x  =  154  ohms  per  wire;  g  =  0.000038  per  wire; 
6  =  0.00107  per  wire;  D  =  10  ft.  =  304.5  cm.  between  wires. 


166  ELECTRICAL  ENGINEERING 

Check  the  constants  of  the  line  and  find  : 

Power  per  phase  at  the  generator  =  P0. 
Current  per  phase  at  the  generator  =  70. 
Voltage  per  phase  at  the  generator  =  E0. 
Volt-amp,  per  phase  at  the  generator  =  E0I0. 

r> 

Power  factor  at  the  generator  =  ^ry 

Power  per  phase  of  the  load  =  P. 

Volt-amp,  per  phase  of  the  load  =  el. 


Power  factor  of  the  load  =  -*• 


p 
-* 

P 


Efficiency  of  transmission 

Voltage  regulation  =  (Eo  —  e)  -f-  e. 

Plot  the  voltage  and  current  vectors  for  both  ends  of  the  line. 
Solution.  —  Resistance  of  No.  000  B.  &  S.  wire,  from  tables, 

=  0.0605o>/1000  ft.  at  60°F. 
.'.  r  =  0.0605  X  5.28  X  200  =  64  ohms. 

Inductance  =  0.00014  logic  ^-^  =  0.00014  loglo  304'5Q5~0'52  per  1000 

ft.,  where  r0  =  radius  =  0.5202  cm. 

.'.  L/1000'  =  0.00014  logio  585  =  0.00014  X  2.767  =  0.0003874. 
L  =  0.0003874  X  5.28  X  200  =  0.409  henry. 
X  =  0.409  X  377  =  154.2  ohms. 


C/1000'  =        '  r  =  0.0074  X  0.361  =  0.002672  micro-farad 

logio—  jr- 

C  =  0.002672  X  5.28  X  200  =  2.82  micro-farads. 
6  =  2T/C  =  377  X  2.82  X  10~6  =  0.00107 

W      corona  loss  per  wire         200,000  nnnnoo, 

9  ~  *    ~  (voltage  to  neutral)  »  =  (72,250)  *  " 
1  The  voltage  to  neutral  on  a  balanced  three-phase  system  is  the  line 

voltage  divided  by  \/3  =  7-=^- 

l.f  O 

The  current  supplied  from  the  generator  is  found  from  (88)  to  be: 

(V7\  I  V  7\ 

1  +  "27  +  €Y  =  (10°  ~  J'50)  I1  +  T)  +  eY- 

YZ  =  gr  -  bx  +j(gx  +  6r);  Y  =  g  -f-  j6;  e  =  72,250. 

/.  -^  =  -  0.081  +;0.037. 

YZ 
1  +-2~  -  0.919  +  J0.037;  eY  =  2.741  +  J77.3. 

Substituting  values, 

I0  =  (100  -j50)  (0.919  +J0.037)  +2.741  +  J77.3 

=  93.75  -  J42.2  -f-  2.741  +  j'77.3  =  96.49  +  J35.1  =  102.5  amp. 


DISTRIBUTED  INDUCTANCE  AND  CAPACITY  167 

The  voltage  at  the  generator  terminals  is  obtained  in  a  similar  way,  and  is 
o  =  e(l  +  ^r)  +  IZ  =  72,250(0.919  +;0.037)  +  (100  - J50)  (64 

=  80,625  +  .;14,928  =  81,200  volts. 

The  power  per  phase  at  the  generator  is,  by  "telescoping"  E0I0, 
Po  =  e0i0  +  e0V  =  80,625  X  96.49  +  14,928  X  35.1  =  8200  kw. 


The  apparent  power  input  to  the  line  is 

#0/0  =  81,200  X  102.5  =  8325  k.v.a.  at  the  generator. 
The  power  factor  at  the  generator  is 

Po        8200 


P.F.o  = 


8325 


0.985. 


The  power  supplied  to  the  load  is 

p  =  ei  =  72,250  X  100  =  7225  kw. 


Load 


FIG.  131. 
The  apparent  power  supplied  to  the  load  is 


el  =  72,250  X  VlOO2  +  502  =  72,250  X  111.9  =  8060  k.v.a. 
The  power  factor  of  the  load  is 

.,'    ;  r.*-£-ig-«"«. 

P        7225 
Efficiency  of  transmission  =  p-   =  OOQA  =  0.882. 


168  ELECTRICAL  ENGINEERING 

81,200  -  72,250 
Regulation   =  -       72  25Q =  12.4  per  cent. 

The  vectors  E0,  70,  e  and  7  are  plotted  to  scale  in  Fig.  130. 
Problem  72. — Consider  a  circuit  as  shown  in  Fig.  131.    Let  the  con- 
stants be: 

r    =  0.01  x    =  0.02 

ri  =  0.01  xi  =  0.02 

r0  =  0.01  x0  =  0.002 

When  the  load  voltage  is  e  =  1,  and  the  load  current  is  7  =  1+  Q.5j, 
find  the  generator  voltage,  current,  power  factor,  and  the  voltage  and  current 
of  the  branch  (r0,  So). 


CHAPTER  XXV 


NOTES    ON    THE   MATHEMATICS    OF    COMPLEX 
QUANTITIES 

This  chapter  is  inserted  in  order  that  the  common  mathe- 
matical operations  shall  be  kept  fresh  in  mind  by  review  and 
frequent  practice.  It  is  very  desirable  that  the  student  shall 
possess  and  retain  facility  in  common,  though  not  always  fre- 
quent, operations.  For  instance: 

Solve  Vo.008,  using  log  tables. 

Solve  6-°-216,  using  log  tables. 

Differentiate  y  =  axn;  y  =  ae~ax; 


y  =  sin  x;  y  =  cos  x; 
u 

y 


u 
uv;  y  =  -• 


Find  the  log,  and  differential,  of  4  —  3j. 
Find  \/-3]. 

Representation  of  Complex  Quantities.— The  general  expres- 
sion for  a  complex  quantity  is  A  =  ai  -f  j«2.     The  numerical 


en 

FIG.  132. 


FIG.  133. 


value,  or  modulus,  of  the  complex  is  A  =  \/ai2  +  a22  and  the 
vectorial  angle  is  tan"1  a  =  — . 

These  various  quantities  may  be  represented  as  in  Fig.  132. 
Then  a:  =  A  cos  a;  a2  =  A  sin  a,  whence  A  =  A  (cos  a  +  j 
sin  Q)  =  Ae3'a,  the  latter  relation  being  proved  later. 

Addition  of  Two  Complex  Quantities.— 

Let  C  =  A  +  B  (Fig.  133). 

169 


170  ELECTRICAL  ENGINEERING 

Then 

C  =  ai  +  ja2  +  61  +  j&2 

=  ai  +  bi  -f  j(«2  +  W> 
and  


V  (ai  +  6])2  +  (as  +  62)2. 


Multiplication  of  Two  Complex  Quantities. 
Let  C  =  A  XB. 

C  =  (ai+ja2)(&i+j&2)' 


and 

C  = 


Division  of  Two  Complex  Quantities.  — 

A 
Let  C  =     - 


4-  j«2 


__  a\l>\ 
and 


Tan  7  = 


Similar   processes   may   be   carried   out   when   the    complex 
quantities  are  expressed  in  polar  coordinates. 

Multiplication.  — 

C  =  AB  =  a(cos  a  +  j  sin  a)6(cos  0  +  j  sin  0) 

=  A£(cos  cy  cos  0  +  j  sin  a  cos  0  +  cos  aj  sin  |S  —  sin  a  sin  ft) 
=  AB(cos  (a  +  0)  +  j  sin(^  +  »). 
Involution  and  Evolution.  — 

A2  =  A  V'a  =  A2(cos  2  a  +  j  sin  2a). 

An  =  An(cos  na  +  j  sin  na). 

A^  =  A^  (cos  -  +  j  sin  -)  (89) 

\       n  nl 


MATHEMATICS  OF  COMPLEX  QUANTITIES     171 


Since  cos  a  =  cos  (a  +  2?rp)  and  sin  a  =  sin  (a  +  2irp)  where 
p  is  any  integer,  the  simple  complex  expression  should  be  written : 

A  =  A  [cos  (a  +  27rp)  +  j  sin  (a  +  2irp)], 

where   there   is  any  question  about  the  number  of  different 
solutions. 

In  evaluating  such  expressions,  a  is  in  radians. 

Sin  X  and  cos  X  may  also  be  written   as   series,1  in  which 


cr 

Sm  #  =  £  — 


r3 
* 


. 

-f-  FT  — 


Cos  x  =  1  -  ry  +  r^  - 


(90) 


Example. — Calculate,    from    series    expression,    the  value  of 
sin  2°.     Since  the  angle  must  be  expressed  in  radians, 

2X27T  7T 

*  =  "360T  =  90  radians' 


Substituting  this  value  into  the  series, 


•    o°  _ 
sm  J     "  90       6  X  903 


=  0.0349. 


120  X  905 

The  Roots  of  a  Complex  Quantity. — Using  the  more  general 
expression,  Eq.  (89)  may  be  written: 

2irp 


cos 


+  j  sin 


(91) 


n  n 

1,  2,  3,  4,  etc.,  and  solve,  continuing 


To  find  the  roots,  put  p 
until  repetition  begins. 

Example. — Find  \/l  = 
where 

A  =  1  =  1  +  jo. 

A  =  A  (cos  a  +  j  sin  a)  =  1 ;  a  =  l',n  =  4;  tan  a  =  y  = 

Tabulating,  and  supplying  values  to  (91) 
p  01234 

2wp 
4 


0 


cos 


0       -1 


0 


-i 


sm 


jO     jl         JO      —jl     JO 

v7A          1     j         -1      -j         1 
The  roots  are  represented  as  vectors  in  Fig.  134. 
1  Developed  by  MACLAURIN'S  theorem. 


-3 

FIG.  134. 


172  ELECTRICAL  ENGINEERING 

Exponential  Representation  of  Complex  Quantities.— The  ex- 
ponent e"  may  be  written  as  a  series  known  as  the  exponential 
series,  developed  from  MACLAURIN'S  theorem. 

Thus, 

u        u2        u3 


Let 

u  =  jB. 
Then, 

*  _  ,    ,    #    ,    £!!  + 

1  +  I  2    +    '    •    '    ' 


je      P  _  JP      ^ 

—     -I-    T~    I  1       '"       O  I    O         I     'I  A 


"(i-f+S)     <« 


These  two  component  series  are  seen  to  be  those  of  the  sine 
and  cosine  (90).     Hence  (92)  may  be  written: 

€j0  =  cos  6  +j  sin  e  (93) 

Since  A  =  A  (cos  a  +  j  sin  a), 

substituting  from  (93), 

A  =  Aeja. 

Thus,  a  third  form  of  writing  the  complex  quantity,  A,  has  been 
developed. 

This  last  may  be  extended  by  letting  A  =  eao.     Thus, 

A   —      «o  Va    _.      ao+jat 
jtl  —   e     c        —   C 

in  which  the  exponent  is  complex. 

Differentiation  of  a  Complex  Number  or  Vector. — 
Let 

A  =  Aeja. 
then 

dA  =  Ajeiada  +  JadA 

'  =  eja[Ajda  +  dA]  (94) 

Logarithm  of  a  Complex  Number  or  Vector.— 

Cdu 
Iqgn-J  — 

We  have 


log  A 


4=    f— 

•-    J  4' 


MATHEMATICS  OF  COMPLEX  QUANTITIES    173 

from  (94), 

rt>"Ajda  .   CdA^     r        rdA 

1  °*A  =  J  ~AfT  +  J  "At*    ==  J  ^  +  J  T 


=  j(a  +  27rp)  +  log  A. 

The  logarithm  of  a  vector  has  thus  an  infinite  number  of 
values. 


/ 
/ 


CHAPTER  XXVI 
THE  TRANSFORMER 

The  alternating-current  transformer  is  used  to  change  electric 
energy  from  one  voltage  to  another.  This  is  done  by  interlinking 
two  electric  circuits  having  different  numbers  of  turns  with  the 
same  magnetic  alternating  flux. 

If  the  two  circuits  enclose  exactly  the  same  flux  it  is  evident 
that  the  voltages  induced  in  the  windings  will  be  proportional  to 
the  numbers  of  turns.  If,  however,  as  is  the  case,  the  flux  is  not 
exactly  the  same  for  each  circuit,  the  ratio  is  slightly  affected  and, 
as  will  be  shown  later,  the  secondary  voltage  has  a  value  differing 
slightly  from  what  the  ratio  of  turns  would  demand. 

When  one  circuit  is  connected  to  an  alternating  e.m.f.,  the 
other  circuit  being  open,  a  current  flows  in  that  circuit  (Fig. 
135).  This  current  is  called  the  no-load  or  the  exciting  current, 
and  may  be  assumed  to  consist  of  two  components,  one  of  which 


Inf 


FIG.  135.  FIG.  136.  FIG.  137. 

supplies  magnetism  to  the  core  and  is  called  the  wattless  com- 
ponent, while  the  other  supplies  power  for  hysteresis  and  eddy 
current  losses  and  is  called  the  power  component. 

These  component  currents  of  the  exciting  current  may  be  rep- 
resented as  flowing  in  a  circuit  of  resistance  and  inductance  in 
parallel  as  in  Fig.  136,  where  e  is  the  e.m.f.  which  sets  up  these 
currents.  They  may  be  represented  vectorially,  as  in  Fig.  137. 
In  the  latter  representation  im,  in  quadrature  with  e,  produces 
the  flux  0,  but  no  power;  4,  in  phase  with  e,  supplies  the  core 

loss.     The  exciting  current,  70o,  lags  behind  e  by  an  angle  tan-1  -r-, 

ih 

It  is  not  strictly  correct  to  represent  the  core  loss  by  a  resist- 
ance r,  Fig.  136,  with  varying  e,  for  part  of  the  core  loss  is  pro- 
portional to  e1-6  and  part  to  e2. 

174 


THE  TRANSFORMER 


175 


Neither  is  it  correct  to  assume  that  the  magnetizing  component 
is  proportional  to  the  e.m.f.,  since  the  magnetization  curve  is  not  a 
straight  line.  However,  in  most  cases,  the  variation  of  e  is 
slight,  and  proportionality  may  be  assumed  without  appreciable 
error. 


The  Transformer  Diagram. — The  relations  of  voltage,  current 
and  flux  which  exist  in  a  transformer  under  normal  operation  are 
shown  with  great  clearness  by  the  aid  of  the  transformer  diagram. 
<f>  represents  the  flux  that  interlinks  with  the  primary  and  second- 
ary of  the  transformer;  et-  is  the  e.m.f. 
induced  in  the  primary  and  secondary 
windings  (assuming  the  same  number 
of  turns  in  each).  This  e.m.f.  is  90° 
in  time  behind  the  flux,  as  is  seen 
from  Fig.  139  and  by  the  following 
simple  proof: 

If 

$  =  <J>m  sin  ( 
then 

N  d<t>  N 


FIG.  139. 


1 2  is  the  secondary  or  load  current  which  in  this  particular 
diagram  is  shown  lagging  behind  the  induced  e.m.f.  72r2  and 
7  2^2  are  respectively  the  e.m.fs.  consumed  by  the  secondary 
resistance  and  reactance,  /2r2  being  in  phase  with  72  and  72z2  being 
90°  ahead  of  72.  72z2  is  the  e.m.f.  consumed  by  the  secondary 
impedance,  which  subtracted  vectorially  from  et  gives  E2  as  the 
secondary  terminal  voltage. 

The  primary  current  may  be  assumed  to  consist  of  three  com- 
ponent parts:  the  first  I\,  which  corresponds  to  the  secondary 
current  and  is  equal  and  opposite  thereto;  the  second  7m,  which  is 


176  ELECTRICAL  ENGINEERING 

the  magnetizing  current  producing  the  flux  <f>  and  is  in  phase 
with  the  flux;  and  the  third  Ih,  which  is  the  power  loss  current  due 
to  the  core  loss  and  is  in  quadrature  to  the  magnetizing  com- 
ponent, that  is,  in  phase  but  opposite  to  the  induced  e.m.f.  et. 

Im  and  Ih  combine  in  700  which  is  the  exciting  current. 

To  overcome  the  induced  e.m.f.  et-  in  the  primary  winding  an 
impressed  e.m.f.  —  e{  is  required. 

To  overcome  the  resistance  and  reactance  drop  in  the  primary 
windings  an  e.m.f.  I\z\  needs  to  be  supplied.  Thus  the  pri- 
mary impressed  e.m.f.  E\  is  the  vector  sum  of  these. 

0i  is  evidently  the  angle  between  the  primary  current  and 
e.m.f. 

62  is  the  angle  between  the  secondary  current  and  e.m.f. 

The  total  primary  current,  /i  =  I'\  -f-  70o. 

In  phase  with  /i  is  the  voltage,  I&1,  consumed  by  the  primary 
resistance,  rj;  and  at  right  angles  ahead  of  /i  is  the  voltage,  I\Xit 
consumed  by  the  self-inductance  of  the  primary  coil. 

These  two  voltages  combine  to  form  /iz1;  the  voltage  consumed 
in  the  primary  of  the  transformer. 

The  total  impressed  primary  voltage,  EI,  is  the  sum  of  IiZi 
and  —  d.  The  angle  0i  is  the  phase  angle  between  EI  and  /i. 

The  transformer  diagram  is  obviously  not  suitable  for  accurate 
calculation.      For  this  purpose,  another  de- 
gj       .     velopment  will  be  made. 

.4,     §j  lf  l        Let  there  be  two  mutually  inductive  coils, 

one  of  them,  called  the  primary,  having  NI 

§L  N     turns,  TI  ohms  resistance,  and  LI  henrys  in- 
ductance, while  the  similar  quantities  of  the 
FIG.  140.  other,  or  secondary  coil,  are  N*y  r2  and  L2 

respectively. 
Then,  in  Fig.  140,  if  the  secondary  current  72  =  0,  the  primary 

impressed  voltage,  ei  =  ilrl  -f  LI  -p  where  e\  and  i\  are  instan- 
taneous values  of  voltage  and  current,  and  Li  is  assumed  con- 
stant. If  a  secondary  current  flows,  there  will  be  induced  in  the 

secondary  an  e.m.f.  et  =  -  L2  -J2-    The  secondary  induced  e.m.f. 

perturn=-f  »£-*?. 

Nt      N2    dt 

If  it  be  assumed  that  there  is  no  leakage,  that  is,  that  all  the 
magnetic  flux  links  with  both  the  primary  and  the  secondary 


THE  TRANSFORMER  177 

coils,  then  the  induced  e.m.f.  in  the  primary  due  to  72  must  be 

Nl  T  di* 
~N~2L^t' 
Then, 


The  sign  of  the  last  term  changes  from  —  to  +  because  the  in- 
duced e.m.f.  must  be  overcome,  or  balanced,  by  an  e.m.f.  of  the 
opposite  sign. 
But 


JV2L2=    \LA 
for,  from  fundamental  relations, 

^1  =  T7^r>  and  #i  = 
Substituting, 

L!  = 

Similarly, 

L2  = 

Then  the  ratio 

Li  _ 
L2~ 


whence 

(96)  then  becomes 


*'r     \/rr 

F2  ^2    Nf^i^2> 


& 


61  =  i\r\  -f~  LI  -37  -f-    \LiL2  j. 
at  at 

Let  \/L]L2  be  denoted  by  3f . 
Then, 

Similarly  for  secondary, 

—   •      _i_  /    ^2  4.   M     l  =  n  (98) 

since  no  e.m.f.  is  impressed  on  the  secondary  coil. 

The  constant,  M,  is  called  the  coefficient  of  mutual  induction, 
and  may  be  defined  as  the  number  of  interlinkages  of  flux  with 
both  coils  of  a  mutually  inductive  circuit  when  unit  current  is 
flowing  in  one  of  the  coils. 

12 


178  ELECTRICAL  ENGINEERING 

Mutual  inductance,  like  self-inductance,  is  measured  in  henrys. 

It  does  not  always  follow  that  M  is  equal  to  \/LiL2.  In  fact, 
that  condition  is  attained  only  when  no  magnetic  leakage  exists, 
which  never  occurs.  If  part  of  the  flux  set  up  by  the  primary 
does  not  interlink  with  the  secondary,  that  part  constitutes  the 
primary  leakage  flux.  Similarly,  when  current  flows  in  the 
secondary,  some  secondary  leakage  flux  is  set  up.  Whenever 
there  is  leakage  flux, 

M 


Eqs.  (97)  and  (98)  hold  at  all  times  provided  the  proper  value  of 
M  is  supplied,  and  M  is  usually  about  95  per  cent,  of  v  LiL2. 

Equivalent  Transformer  Circuit.  —  The  differential  equations 
given  above  are  not  readily  used,  but,  fortunately,  STEINMETZ 
has  evolved  a  simple  treatment  involving  a  diagram  of  simple 
series  and  multiple  circuits,  which,  while  not  showing  the  physics 
of  the  phenomenon,  lends  itself  to  very  simple  and  quite  accurate 
treatment. 

He  represents  the  transformer  by  a  circuit  which  is  shown  in 
Fig.  141. 


rl 

% 

ra 

Ja 

t 

^     t      ri. 

f« 

*oo| 

Si 

l&oo 

»M 

1 

I 

1 

i  V 

FIG.  141. 

Let  the  secondary  or  load  current  be  7  2  =  iz  +  ji't,  and  let 
the  secondary  terminal  voltage  be  e2,  the  zero  vector. 

Then  the  secondary  induced  e.m.f  .  Et  =  e*  +  hZ2  =  ez  +  itfz  — 


The  exciting  current  is 

/oo  =  EiY0Q  =  (€i  -f  je'i)  (goo  +  j&oo) 


The  primary  current  is 

/i   =   /2  +  /oo  =  ^2  +  IO 

The  impressed  voltage  is 


THE  TRANSFORMER  179 

From  these  values  may  be  obtained: 

power  output  =  e^iz, 

power  input  =  &iii  -f  e\i'\t 

f  .  .     ,         power  input       e\i\  +  d\i\ 

power  factor  at  primary  terminals   =  — — rr-      —  = '  T       > 

volt-amp.  EJi 

regulation  = 


€2 
62*2 


efficiency 


In  using  these  equations  r*i,  r2,  Xi  and  x2  are  positive,  &0o  is 
negative  because  the  magnetizing  circuit  is  necessarily  inductive, 
iz  is  negative  for  lagging,  positive  for  leading  current. 

Transformers  are  rated  on  the  basis  of  kilovolt-amperes,  not 
kilowatts. 

Example  of  Transformer  Calculation. — Given  a  2200  to  220- 
volt,  60-cycle,  50-kv.a.  transformer,  in  which  r*i  =  0.97,  r2  = 
0.0097.  Assume  that  on  test  98.5  volts  on  the  primary  produces 
full-load  current  in  the  short- 
circuited  secondary  (142,  a). 
At  no-load,  with  the  normal 
voltage  (220)  impressed  on  the 
secondary,  the  primary  circuit 
being  open,  the  watts  input  are 
WQ  =  1000,  and  the  exciting 
current  /Oo  =  12.25  amp.  (Fig. 
142,  6).  The  percentage  rl 
drop  in  the  primary  is 


22.7 


(6) 
FIG.  142. 


22.7  X  0.97 
-2200"      =  0.01  ==  1  per  cent. 

where  22.7  is  the  normal  primary  current. 


In  the  secondary, 
per  cent,  rl  drop  = 


=  0.01  =  1  per  cent. 


The  total  impedance,  calculated  from  the  short-circuit  test 
(142,  a),  is 

QO    K 


ZMal  =  2277 


4.35  ohms. 


180  ELECTRICAL  ENGINEERING 

Total   per   cent,  impedance  drop,  referred   to   the  primary 
voltage,  is 


=  °-0448  =  4'48  Per  cent' 


<n 
2200 

Percentage  total  reactance  drop  is  then  V4.482  —  22  =  4  per  cent. 
total  per  cent,  resistance  drop  =  1  +  1  =  2). 

Therefore,  assuming  primary  and  secondary  percentage  react- 
ances to  be  equal, 

per  cent.  Xi  =  2  per  cent.;  per  cent.  xz  =  2  per  cent. 
Thus,  on  the  percentage  basis,  or  assuming  e2  =  1  and  i  =  1,  then, 

ri  =  0.01  xi  =  0.02 

r2  =  0.01  xz  =  0.02. 

The  core  loss  of  1000  watts  obtained  on  test  is  supplied  at  220 
volts  by  the  component  of  no-load  current,  4. 

1000 

4.55  amp. 


The  per  cent.  4  of  the  secondary  current  is 

per  cent.  ih  =        -  =  0.02  =  2  per  cent. 


0.02. 


The  magnetizing  component  of  the  no-load  current  is  obtained 
from  7oo  and  the  core-loss  current.     Thus, 


V  12.252  -  4.552  =  11.35  amp. 

11  35 

The  percentage  im  =      '„    =  0.05  =  5  per  cent. 

=  -  0.05. 


Having  obtained  the  above  constants,  values  may  now  be  tabu- 
lated to  find  the  effect  of  variation  of  the  load  current  with 
constant  power  factor. 

Problem  72. — Let  power  factors  of  100  per  cent.,  80  per  cent,  lagging  and 
80  per  cent,  leading  be  assumed,  and  let  the  calculations  be  made  for 
secondary  currents  of  0,  0.5,  and  1. 


THE  TRANSFORMER 


181 


Tabulating : 


0.8  Lagging 


P.F.  =  unity 


0.8  Leading 


12 .. 
i't. 


oo. . 
e'  iff o 


t  oo. . 
ii. . . . 


ei 

t'm.. 


Pj. . . 
em . . 
e'li'i. 
Pi... 


P.F.i. 
Eff... 
Reg.. 


0.0 
0.0 
0.0 
0.0 
0.0 

0.0 
0.0 
1.0 
0.0 
0.02 

0.0 

0.02 

0.0 
-0.05 
-0.05 

0.02 

-0.05 
0.054 
0.0002 

-0.001 

1.0012 
-0.0005 

0.0004 
-0.0001 

1.001 

0.0 

0.02 

0.0 

0.02 

0.054 

0.37 

0.0 

0.001 


—0.00025—0 
0.02045 
0.0001 
0.0505 
0.0504 


0.5 

0.4 
-0.3 

0.004 
-0.006 

0.008 
-0.003 
1.01 
0.005 
0.0202 


0.42045 
-0.3504 
0.547 
0.0042 
-0.007 

1.0212 
-0.0035 
0.0084 
0.01 
1.021 

0.4 
0.49 
-0.003 
0.426 
0.557 

0.762 
0.939 
0.021 


1.0 

0.8 
-0.6 

0.008 
-0.012 

0.016 
-0.006 
1.02 
0.01 
0.0204 


.0005 
0.0209 
0.0002 
-0.051 
-0.0508 


0.8209 

-0.6508 

1.046 

0.0082 

-0.013 

1.0295 
-0.0065 
0.0164 
0.02 
1.03 

0.8 
0.854 
-0.013 
0.841 
1.087 

0.772 
0.951 
0.03 


0.0 
0.0 
0.0 
0.0 
0.0 

0.0 
0.0 
1.0 
0.0 
0.02 

0.0 

0.02 

0.0 
-0.05 
-0.05 

0.02 

-0.05 
0.0538 
0.0002 

-0.001 

1.0012 
-0.0005 

0.0004 
-0.0001 

1.002 

0.0 

0.02 

0.000001 

0.02 

0.0538 

0.372 
0.0 

0.002 


0.5 

0.5 

0.0 

0.005 

0.0 

0.01 

0.0 

1.005 

0.01 

0.0201 

-0.0005 

0.0206 

0.0002 

-0.0505 

-0.05005 

0.5206 

-0.05005 

0.522 

0.0052 

-0.0010 

1.011 
-0.0005 
0.0104 


1.012 

0.5 
0.5265 
-0.001 
0.525 
0.5272 

0.995 
0.952 
0.012 


1.0 
1.0 
0.0 
0.01 
0.0 

0.02 

0.0 

1.01 

0.02 

0.0202 

-0.001 
0.0212 
0.0004 
-0.0505 
-0.0501 


0.0 
0.0 
0.0 
0.0 
0.0 

0.0 
0.0 

1.1 

0.0 
0.02 

0.0 

0.02 

0.0 

-0.05 

-0.05 


1.0212 
0.0501 
1.022 
0.0102 
-0.0010  -0 


0.02 
-0.05 
0.0538 
0.0002 
001 


1.021 
-0.0005 


0.0403 
1.022 

1.0 
1.041 
-0.002 
1.039 
1.0404 

0.999 
0.963 
0.022 


1.0012 
-0.0005 

0.0004 
-0.0001 

1.0013 

0.0 

0.02 

0.000001 

0.02 

0.0538 

0.372 

0.0 

0.0013 


0.5 

0.4 

0.3 

0.004 

0.006 

0.008 
0.003 
0.998 
0.011 
0.01996 

-0. 00055 ; 

0. 02051 ! 

0.00022 
-0.0499 
-0.0497 

0.4205 
0.2503 
0.4965 
0.0042 
0.005 

0.997 

0.0025 

0.0084 

0.0219 

0.9974 

0.4 

0.419 

0.0055 

0.424 

0.494 

0.859 

0.942 

-0.0026 


1.0 
0.8 
0.6 


0.012 

0.016 
0.006 
0.996 
0.022 
0.01992 

-0.0011 
0.02102 
0.00044 
-0.0498 
-0.0494 

0.82102 

0.55U6 

0.988 


0.011 


0.0055 
0.0168 
0.0443 
0.9932 

0.8 

0.815 

0.0242 

0.839 

0.975 

0.86 
0.954 


Problem  73. — Write  a  discussion  of  the  results  obtained  in  problem  72  for 
the  three  values  of  current  and  the  three  power  factors. 

Approximate  Method  of  Determining  the  Regulation,  Effi- 
ciency and  Power  Factor  of  Transformers. — Let  1 2  =  iz  -f-  ji\  be 
the  secondary  current. 

Then  the  primary  current  is  approximately 

/i  =  72  +  4  —  Jim  =  iz  +  ih  +  j(ir*  —  im)  =  ii  +  ji'i- 
In  the  secondary  winding  only  the  secondary  current,  I 2,  flows; 
in  the  primary  winding,  the  primary  current.     The  average  cur- 
rent in  the  two  windings  considered  as  one  is,  then, 

la    =    12   +   0.54   +j (^2   ~   0.54). 

If  the  secondary  voltage,  referred  to  primary,  is  the  zero  vector 
and  is  e2,  then  the  primary  voltage  is 

#1    =    £2   +  /«Zo, 


182  ELECTRICAL  ENGINEERING 

where  Z0  =  r0  +  jx*  is  the  sum  of  the  impedances  of  the  primary 
and  secondary  referred  to  the  primary. 

Ei  =  62  +  [i%  +  0.54  +  j(*'»  -  O-5*"*)!  (r°  +  ^o) 
0.54r0  -  *Vo  +  O.S^zo 


and  the  real  value  of  the  primary  voltage  is,  neglecting  second 
power  of  small  terms,         _  __ 
El  =  vV  +  2e2fero  +  0.5ur0  -  i'*x*  +  0.5imxo) 
Regulation  is 

^~62  =  ^1  _  l  (100) 

62  02 

Let  Ei  and  /i  represent  the  primary 
e.m.f.,  and  current,  as  in  Fig.  143. 
Then  the  power  input  is 


P  = 


cos  B. 


But 


cos  6  =  cos  (a  -  j3)  =  cos  a  cos  0  +  sin  a  sin  /3 


:.  p 


x 


The  secondary  output  is  62^2- 
The  primary  input  is  e\i\  +  e'\if\, 


=  e2i2  +  ^4  +  I^TO,  approximately. 
The  efficiency  is 

Output  #2^2 

input        £2^2  +  e^ik  +  I  fro 
,        Similarly, 

pi  =  -  e2i'2  +  e2im  +  hzxQ,  approximately. 


(101) 


and  cos  0i  is  the  power  factor  of  the  primary. 

Problem  74.— (A)  Determine  the  numerical  values  of  the  primary  and 
secondary  resistances  and  reactances,  the  core-loss  current,  the  magnetizing 
current,  the  exciting  current  from  the  1000-volt  side,  the  core  loss  in  watts, 


THE  TRANSFORMER 


183 


and  the  short-circuit  impedance  when  taken  from  the  1000-volt  side,  for  the 
following,  1000-100  volt  transformers.  The  primary  and  secondary  re- 
sistance drops  are  each  1  per  cent. ;  the  primary  and  secondary  reactance 
drops  are  each  2  per  cent. 

The  conductances,  00o,  and  susceptances,  60o,  are  calculated  at  1000  volts. 


Rating  in  k.v.a. 

10 

20 

40 

80 

160 

320 

0.0002 
0.0006 

0.0004 
0.0012 

0.0008 
0.0024 

0.0016 
0.0048 

0.0032 
0.0096 

0.0064 
0.0192 

Problem  74.  —  (B)  Find  the  power  factor,  regulation  and  efficiency  of  these 
transformers  by  the  approximate  method,  assuming  unity  power  factor  of 
load. 

Problem  74.  —  (C)  For  any  one  transformer,  plot  the  regulation  and  effi- 
ciency vs.  power  factor,  and  find  the  points  of  0  per  cent,  regulation  and 
maximum  efficiency. 

(A)  Solution  for  the  10  k.v.a.,  1000-100  Volt  Transformer.—  Since,  with  non- 
inductive  load  (on  the  secondary)  the  primary  voltage  and  current  will  be 
nearly  in  phase,  the  approximate  primary  current  is 
10,000  k.v.a. 
" 


/1= 


1000  volts 


10amP' 


rj  drop  =  1  per  cent.  =  0.01  X  1000  volts  =  10  volts 

10  volts 

TI  =  T^~~ '  —  1  ohm. 

10  amp. 

TZ\ "  =  0.01  ohm. 


Similarly, 

Xi  =  2  ohms. 
X2  =  0.02  ohm. 
The  core-loss  current  is 

ih  =  eg00  =  1000  X  0.0002  =  0.2  amp. 
Magnetizing  current  is 

im  =  eb00  =  1000  X  0.0006  =  0.6  amp.,  lagging, 
.'."no-load  current  is 

/oo  =  \/ihz  +  im2  =  \/0.04  +  0.36  =  0.632  amp. 
The  core  loss  is 

eVoo  =  10002  X  0.0002  =  200  watts. 
The  short-circuit  impedance  is 


=  \/22  +  42  =  -\/2Q  =  4.47  ohms. 

(B)  Solution— E!  =  1000  =  V^2  +  2e2(i2r0  +  0.5^r0  +  0.5  imx0) 
Tabulating  for  equations  (100),  (101),  (102): 

10.0  erfi  9820.0 


Kw. 


72 


10.0 
1.0 


196.4 
100.0 


184 


ELECTRICAL  ENGINEERING 


AiAi 
U) 


0.5  t 

im 

0.5  i 
e2 
JSi, 

e2 
Reg. 


1.0 

2.0 
2.0 

2.0 

4.0 
20.0 

0.2 

0.2 
-0.6 
-1.2 
982.0 

1.018 
0.018 


»Vr0 

Eff. 


tan  <p 
P.F.  =  cos  0 


200.0 

0.961 
400.0 

-589.2 

-189.2 
10,216.4 
-0.01853 
0.9998 


(C)  Solution. — In  finding  the  efficiency  and  regulation  it  makes  no  differ- 
ence in  the  results  whether  the  problem  is  solved  on  the  percentage  basis  or 
by  supplying  numerical  values  for  any  given  machine. 

The  former  method  is  more  general  in  its  application,  and  it  will  be  used 
here,  percentage  values  being  taken  from  the  data  of  the  10-kw.  transformer 
and  applied  in  formulae  (100)  and  (101). 
The  percentage  data  then,  are : 

El  =  1,  /2  =  1,  ih  =  0.02,  im  =  -0.06,  r0  =  0.02,  x0  =  0.04 


0      0.25    0.50  0.75    1.00   0.75    0.50  0.25 
Lagging  Leading 

Power  Factor 

FlG.  144. 


THE  TRANSFORMER 


185 


000000000   05   000000 
OOr-  (OOOOOi-i   O   OOOOOO 

III  1 


*O   00    O 

8  3  g 


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III  1 


OO 


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OOO 


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III  1 


OOO 


OOOOOOOT-H      O      OOOOOO 
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IOIOO'-HO<MOI^»O        Tt<        T^rHrH 

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1  1 


lOtNt^CllN  lOrH  CO 

CD   (M    O   T^    i—  t    CO    »O  N.  l>t^Oi           CO1^ 

COrHOCOO^»O  T^  T^t^i-tC<Ji-l(M 

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OOOOOOOOO  I-H  OOOOOO 

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W   00 
rH<MGO      •* 


00 
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OOOOOOOOO  1-1  OOOOOO 

I I 

<M          <N  (N          <N 

CO          i— ^    O5   C^  O5  O5          O5          O5 

OOi— IOOOOOO5  O  OOOOOO 

ooooooooo  i-l  doodod 

I     I 


S.3 


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•  v  10  •?  10   . 
;^o'o 


bO 

£ 


b£ 


1 

>> 

I 

o 

£ 

0> 


w 


CHAPTER  XXVII 
HYSTERESIS  AND  EDDY  CURRENT  LOSSES 

Hysteresis  Loss. — The  hysteresis  loop  is  interesting  in  that  it 
indicates  by  its  area  directly  the  work  done  on  the  electromagnet 
per  cycle  of  change  of  current. 

The  work  done  in  an  electric  circuit  has  been  shown  to  be 
feidt.  If  T  is  the  time  of  the  cyclic  variation  of  current  then, 

CT 
W  =    I  eidt,  is  the  work  performed  during  the  cycle. 

JQ 
But  the  induced  e.m.f.  in  a  winding  of  N  turns  is 

e  =  —  jgg  -=r>  where  -^  is  the  rate  of  change  of  flux. 

.'.  W  =  —  I  TT^  -j7  dt.  But  (j)  =  SB,  where  S  is  the  cross- 
sectional  area  of  the  magnetic  circuit  in  square  centimeters. 

.  W.  .  rx&Ma 

Jo    108    dt  C 
Also  the  magnetizing  force  is: 

QAiriN 
tf  =  -_— , 

where  i  is  given  in  amperes. 
Thus, 

•A7  IH 

iN  = 


0.47T' 

and,  substituting  this  value, 

r   SIH    as  si    r 

J0  0.47T  xW* ~di c        '  WxT*h  HdB' 

But  SI  is  the  volume,  7,  of  the  magnetic  structure.     Thus, 

V        CT 

W  =  -  HdB. 

107  X  47rJ0 

But  HdB  is  the  area  of  the  hysteresis  loop  corresponding  to 
maximum  density,  B,  as  seen  from  the  loop.  The  work  is  given 
in  joules. 

186 


HYSTERESIS  AND  EDDY  CURRENT  LOSSES    187 

STEINMETZ  found  that  the  hysteresis  loss  in  watts  could  be 
expressed  (approximately)  by  the  following  equation: 

W 


107 

where  V  is  the  volume  and  rj  is  a  constant  which  depends  upon 
the  quality  of  the  iron.  The  equation  shows  that  the  loss  is 
proportional  to  the  1.6  power  of  the  maximum  density  and 
directly  proportional  to  the  frequency.  In  centimeter  measure- 
ments ri  varies  from  0.001  to  0.002  in  ordinary  sheet  iron  and 
may  be  10  times  as  great  in  tempered  steel.  In  the  best  silicon 
steel  it  is  0.0006,  which  corresponds  to  0.54  watt  per  Ib.  at  60 
cycles  and  a  density  of  64,500  lines  per  sq.  in.  or  10,000  lines 
per  sq.  cm. 

Eddy  Current  Loss. — Eddy  currents  differ  in  no  way  from  other 
currents,  and  the  loss  of  power  by  them  is  therefore  i2R  or  if  E 
is  the  e.m.f.  causing  the  current  and  Z  is  the  impedance  of  the 
path,  then, 


and  the  loss  is 


It  follows,  then,  that  the  loss  is  proportional  to  the  square  of  the 
e.m.f.  or,  what  is  equivalent,  to  the  square  of  the  maximum 
density  and  to  the  square  of  the  frequency,  since  the  e.m.f. 
itself  is  proportional  to  the  frequency  of  flux  variation  and  the 
maximum  density. 

Even  in  the  simplest  cases  it  is  difficult  to  calculate  the  loss 
since  the  distribution  of  the  flux  and,  therefore,  the  e.m.f.  in 
different  parts  of  the  material  is  often  very  complex. 

Consider  as  an  illustration  the  simple  case  of  eddy  current 
loss  in  transformer  steel.      The  cores  are 
built  up  of  laminations  in  such  a  way  that 
the  flux  path  is  divided  up  into  a  number 
of  elements  each  having  the  section  of  the 
edge  of  a  lamination  and  following  parallel, 
or  as  nearly  so  as  possible,  to  the  sides  of          "^j^   145 
the  laminations. 

With  the  flux  entering,  as  is  shown  in  Fig.  145,  currents  will 
flow  as  indicated  by  the  dotted  lines.  The  current  flowing 


188  ELECTRICAL  ENGINEERING 

through  a  section  of  area  l\dx  encloses  a  flux  which  is  —  r  0,  where 

u 

</>  is  the  flux  passing  through  the  entire  area  of  one  lamination 
(assuming  uniform  flux  density).1 
The  effective  value  of  the  e.m.f.  induced  is 

4.44  X  flux  X  turns  X  frequency  _    \/27r2o;<ft/ 
108 


The  resistance  of  the  path,  neglecting  that  of  the  ends,  is 

2lp 

lidx 

.        \/2ir2x<j>flidx 


where  p  is  the  specific  resistance  of  the  material. 
.'.  t'2r  in  the  elementary  circuit  is 

x    2/p          4 


p2  h(dx) 

and  the  total  loss  is 


p  '  4**<l>*f*l0*(dx) 
Jo          IVWp 


6  X  1016ZP 
Since  the  volume  is  Hid,  the  loss  per  cm.3  is 

W  TT^hd  1  TT22 

X 


F        6  X  1016ZP       Z  X  W  ~  6  X  1016/2p 
But  0  =  5  X  W.     /.  </>2  = 

and 

W_        Tr2BH2d2f* 

V    ~  6  X  1016Z2p  ~  6  X  10"p  Watts* 

p  for  sheet  iron  is  about  ^  ohms. 

1  For  a  more  complete  discussion  see  "  Advanced  Electrical  Engineer- 
ing." 


CHAPTER  XXVIII 
WAVE  DISTORTION  IN  TRANSFORMERS 

If  on  a  transformer  containing  no  iron  a  sine  wave  of  e.m.f. 
were  impressed  at  its  terminals,  the  flux  and  the  exciting  current 
would  also  follow  sine  waves. 

With  the  introduction  of  iron,  however,  while  the  flux  values 
would  still  follow  a  sine  wave,  or  very  nearly  so  —  being  distorted 
only  due  to  the  ohmic  drop  of  the  distorted  current  —  the  exciting 
current  wave  would  necessarily  be  considerably  distorted. 

Its  shape  is  shown  in  Fig.  148,  which  is  derived  from  the  hyste- 
resis loop  given  in  Fig.  147. 

Conversely,  if  by  some  arrangement  the  exciting  current  were 
made  to  follow  substantially  a  sine  wave,  the  flux  wave,  and 
therefore  the  wave  of  voltage  across  the  transformer,  would  be 
greatly  distorted. 

This  distortion  in  current  or  e.m.f.  waves  is  of  considerable 
importance  in  connection  with  the  grouping  of  transformers  in  a 
three-phase  system,  as  will  be  seen  later.  At  present,  however, 
only  the  condition  in  a  single-phase  transformer  will  be  studied. 

A  representative  hysteresis  loop  is  shown  in  Fig.  147,  which  was 
obtained  from  actual  tests  with  a  sine  wave  of  impressed  e.m.f. 
The  test  data  are  recorded  in  Table  VI. 

If  the  effect  of  the  ohmic  drop  be  neglected,  then  the  impressed 
and  counter,  or  induced,  e.m.f.  are  the  same  numerically  and 


where  N  is  the  number  of  turns  and  <£  is  the  flux. 
With  a  sine  wave  of  flux  </>  =  $>m  sin  co£, 

dt 

-^  =  4>TOco  cos  at. 

cos  a)t  =  —  Em  cos 


The  induced  e.m.f.  has  its  negative  maximum  when  the  flux 
begins  to  rise,  and  lags  behind  the  flux  by  90  time  degrees.     Thus 

189 


190 


ELECTRICAL  ENGINEERING 


the  impressed  e.m.f  .,  E,  which  is  equal  and  opposite  to  the  induced 
e.m.f.,  leads  the  flux  by  90°  (neglecting  the  ir  drop),  Fig.  146. 

If  instead  of  being  a  sine  wave  the  flux  were  distorted  and  yet 
symmetrical,  it  would  be  represented  by  FOURIER'S  series  of  odd 
harmonics,  thus: 

0  =  3>lm  sin  at  +  <J>3wi  sin  (3o>Z  +  a) 

/.  e.  =  —  N  -ir  =  —  &imw  cos  ut 

The  e.m.f.  wave  would  be  relatively 
more  distorted  than  the  flux  wave  as 
is  evident  from  the  coefficients  of  the 
different  trigonometric  terms. 


sn 


cos 


-fa)—... 


FIG.  147. 


When  a  hysteresis  loop  is  given,  if  either  the  flux  wave  or  ex- 
citing current  wave  is  known,  the  other  may  be  at  once  obtained. 
For  example,  let  the  flux  wave  be  assumed  to  be  sinusoidal. 
TABLE  VI. — HYSTERESIS  LOOP  DATA 


Ord. 

Abs.                                          Aba. 

0.0 

0.5 

-0.5 

0.2 

0.56 

-0.43 

0.4 

0.63 

-0.32 

0.6 

0.71 

-0.18 

0.8 

0.82 

0.08 

0.9 

0.9 

0.35 

1.0 

1.0 

1.0 

EXCITING  CURRENT  DATA 

Time 

Flux 

ioo 

0 

0.0 

0.5 

10° 

0.174 

0.55 

20° 

0.34 

0.6 

WAVE  DISTORTION  IN  TRANSFORMERS        191 

The  exciting  current  data  are  obtained  from  the  hysteresis  loop 
by  reading  off  the  current  values  corresponding  to  the  flux  values 
which  have  been  taken  at  uniform  intervals  along  the  flux  wave. 
Thus,  at  0°  on  the  flux  wave  $  =  0.  This  value  of  <£,  on  the 
hysteresis  loop,  corresponds  to  i00  =  0.5  amp.  At  10°  on  the 
flux  wave,  <£  =  0.174.  This  value  on  the  loop  corresponds  to 
z'oo  =  0.55,  etc.  Data  for  the  exciting  current  are  given  in  Table 

G/T.  It  should  be  noted  that  the  flux 
naximum  and  current  maximum  always 
>ccur  at  the  same  instant.  /  *°°- 

The  phase  relations  and  character- 
istic current  wave  shape  for  a  sine  wave 
of  flux  are  shown  in  Fig.  148  The  im  FIG 

pressed  voltage  wave  leads  the  flux  by 

90°.  The  scales  to  which  the  waves  are  plotted  are  quite  in- 
dependent of  each  other,  and  should  be  so  chosen  as  to  exhibit 
the  waves  most  clearly. 

When  the  induced  e.m.f.  is  not  a  sine  wave,  the  flux  wave  is 
also  distorted.  In  this  case  the  impressed  e.m.f. 

-N**. 

~  *  dt 

Transposing, 

edt 

where  N  is  the  number  of  turns. 
Hence 


ft   -   <2  />2 

I  «&•-       A^d0. 

./«  =  «i       ^i 


If  ti  is  chosen  as  the  time  when  <f>  is  zero,  and  tz  is  the  time  when 
</>  is  maximum,  then 


AC-*         ^N  N 

!«*-       ioi^  =  ioi 

Ji  =  (i         t/0 


This  equation  shows  that  the  maximum  value  of  the  magnetic 
flux  or  flux  density— in  which  the  electrical  engineer  is  very  much 
interested,  since  it  determines  the  magnetizing  current  and  core 
loss— is  proportional  to  a  certain  area  of  the  e.m.f.  wave,  and  it 
remains  to  determine  where  this  area  is  located. 

When  the  flux  is  a  maximum  then  -—  is  zero;  thus  e  is  zero. 


192 


ELECTRICAL  ENGINEERING 


The  value  of  tz  is  therefore  easily  ascertained,  as  is  shown  in  Fig. 
149. 

The  ordinate  through  ti  must  bisect  the  e.m.f.  wave  in  order 
that  the  flux  wave  be  symmetrical,  as 
can  also  be  seen  by  slight  consideration, 
since  the  flux  wave  must  be  symmetrical 
above  and  below  the  zero  line. 

Thus,  in  finding  the  flux  wave,  the 
first  step  is  to  bisect  the  area  of  the 
e.m.f.  half-wave,  which  gives  the  posi- 


FIG.  149. 


tion  of  ti  and  the  zero  of  the  flux  wave. 

Problem  76. — From  the  following  readings  on  a  distorted  e.m.f.  wave 
obtain  and  plot  the  flux  and  current  waves. 
NOTE. — Choose  a  scale  to  give  <bm  —  1. 


t 

ei 

t 

Ci 

0° 

0.0 

100° 

0.73 

10° 

0.005 

110° 

0.90 

20° 

0.01 

120° 

1.0 

30° 

0.04 

130° 

0.98 

40° 

0.1 

140° 

0.91 

50° 

0.15 

150° 

0.78 

60° 

0.22 

160° 

0.5 

70° 

0.31 

170° 

0.12 

80° 

0.42 

180° 

0.0 

90° 

0.58 

Solution. — By  bisecting  the  area  of  the  e.m.f.  half-wave  it  is 
found  that  the  zero  of  the  flux  wave  will  be  at  120°  in  this  ex- 
ample. This  is  also  the  point  of  maximum  e.m.f.  Starting 
from  120°  and  tabulating  values  proportional  to  the  areas 
enclosed  for  each  10°  gives  values  proportional  to  the  flux  when 
these  areas  are  successively  summed  up.  Thus  at  120°,  flux  =  0. 
At  130°,  the  area  enclosed  between  120°  and  130°  ordinates  and 
the  curve  and  base  line  is  proportional  to  the  mean  ordinate,  say 

— 2~~    ~  =  0.99.     At  140°,  the  mean  ordinate  between  130° 
and  140°  is  0.95. 

The  area  from  120°  to  140°  is  proportional  to  0.99  +  0.95  = 
1.94.  Thus,  three  points  on  the  curve  are  obtained,  namely, 
0,  0.99,  1.94. 


WAVE  DISTORTION  IN  TRANSFORMERS        193 


These  values  may  conveniently  be  reduced  by  a  factor  to 
bring  the  maximum  of  the  flux  wave  to  unity. 
The  tabulation  is  as  follows : 


t 

120° 

130° 

140° 

150° 

160° 

170° 

180° 

€\ 

1.00 

0.98 

0  91 

0  78 

0  50 

0  12 

0  0 

Av 

0  99 

0  95 

0  85 

0  64 

0  31 

0  06 

Area       

0.0 

0.99 

1.94 

2  79 

3  43 

3  74 

3  8 

0  263  X  area 

0  0 

0  26 

0  51 

0  735 

0  903 

0  985 

1  00 

t 

190° 

200° 

210° 

220° 

230°  \ 

) 
240° 

€i                      

-0.005 

-0.01 

-0.04 

-0.10 

-0.15 

-0.22 

Av  

-0.0025 

-0.0075 

-0.025 

-0.07 

-0.125 

-0.185 

Area  

3.8 

3.79 

3.77 

3.7 

3.57 

3.39 

0.263  X  area.  . 

1.00 

0.997 

0.992 

0.975 

0.94 

0.893 

t 

250° 

260° 

270° 

280° 

290° 

300° 

€i  

-0.31 

-0.42 

-0.58 

-0.73 

-0.90 

-1.00 

Av  

-0.265 

-0.365 

-0.50 

-0.655 

-0.815 

-0.95 

Area  

3.12 

2.76 

2.26 

1.6 

0.785 

-0.165 

0.263  X  area.  . 

0.822 

0.727 

0.595 

0.421 

0.206 

0.0 

1.0 
0.8 
0.6 
0.4 
0.2 
0 
-0.2 
-0.4 
-0.6 
-0.8 
-1.0 

f 

N 

\    / 

^ 

/ 

/\ 

«• 

/ 

ty 

;       * 

« 

y 

\ 

^ 

S 

/ 

V. 

0 

4 

0 

8 

0 

/ 

20 

1 

0 

2( 

0~*^s 

/ 

. 

/ 

^ 

X 

Angular  Displacement 

FIG.  150. 


13 


194  ELECTRICAL  ENGINEERING 

The  tabulation  is  carried  out  for  values  of  et  from  120°  to 
300°,  values  from  180°  to  300°  being  the  same  as  from  0°  to  120° 
but  reversed  in  sign. 

The  flux  wave  is  then  plotted  from  0°  to  120°  by  reversing  the 
sign  of  the  values  of  flux  obtained  from  180°  to  300°.  These 
waves  are  shown  in  Fig.  150. 

Problem  76. — With  three-phase  systems,  the  exciting  current  of  Y-con- 
nected  transformers  resembles  fairly  closely  a  sine  wave.1  Assuming, 


FIG.  151. 

therefore,  a  sine  wave  of  exciting  current,  determine  the  flux  wave  from  the 
hysteresis  loop  (Fig.  147),  and  from  this  find  and  plot  c,-.  These  waves 
are  shown  in  Fig.  151,  in  which  the  characteristic  form  of  the  induced  vol- 
tage, 6i,  is  noteworthy. 

Problem  77. — Analyze,  by  FOURIER'S  series,2  the  typical  wave  of  exciting 
current  shown  in  Fig.  148,  determining  and  plotting  the  fundamental  and 
third  harmonic  and,  if  sufficient  time  is  available,  also  the  fifth  harmonic. 

Dependence  of  Core  Loss  on  the  Shape  of  the  E.M.F.  Wave. — 

The  core  loss  of  a  transformer,  which  is  due  to  hysteresis  and 
eddy  currents  in  the  iron  core  and  is  equal  to  e»Voo,  depends  on  the 
maximum  value  of  the  flux,  since  the  greater  the  maximum  flux 
the  greater  the  area  enclosed  by  the  hysteresis  loop.  In  modern 
transformers,  hysteresis  loss  is  about  70  per  cent,  and  eddy  current 
loss  about  30  per  cent,  of  the  core  loss. 

But  $m  depends  upon  the  area  of  the  e.m.f.  wave,  as  has  been 
illustrated  in  the  problems,  and  hence  on  the  average  value  of 
the  e.m.f. 

Hysteresis  loss  is  approximately  proportional  to  the  1.6th 
power  of  the  maximum  flux. 

Thus,  if  a  comparison  is  made  of  two  e.m.f.  waves  of  equal 
effective  value,  but  of  different  shape  and  average  value,  the 
ratio 

Hysteresis  loss  in  wave  A  _  /av.  e.m.f.  of  A\  1>6 
Hysteresis  loss  in  wave  B  ~  \av.  e.m.f.  of  B/ 

1  This  is  demonstrated  on  p.  228,  Chap.  XXXII. 
8  See  Chap.  XXIX. 


WAVE  DISTORTION  IN  TRANSFORMERS        195 


By  definition, 


Form  factor  (f .f.)  = 


effective  e.m.f. 
average   e.m.f. 


.  Hysteresis  loss  in  A  _  rf.f.  (B)"| L6 
'  'Hysteresis  loss  in  B      Lf.f.  (A)J 


Therefore,  the  higher  the  form  factor  the  less  the  core  loss. 
The  form  factor  of  a  sine  wave  is  1.1.  In  general  that  of  a  flat- 
top wave  is  less;  of  a  peaked  wave,  more. 

Wave  A  (Fig.  152)  has  maximum  core  loss. 

Wave  B  has  minimum  core  loss. 


FIG.  152. 


CHAPTER  XXIX 


DISTORTED  WAVES 

It  is  often  necessary  to  express  a  distorted  wave  in  the  form 
of  an  equation.  This  can  readily  be  done  since  it  has  been 
found  that  any  periodic  univalent  curve  can  be  expressed  by  a 
series  of  terms  involving  a  constant  and  sine  and  cosine  terms. 
That  is, 
y  =  a0  +  ai  cos  0  +  «2  cos  26  + +  an  cos  nB 

+  61  sin  0  +  62  sin  20  + +  6nsinr*0  (103) 

represents  any  distorted  wave  in  which  for  every  value  of 
abscissa  only  one  ordinate  exists,  provided  that  the  abscissa 
is  so  chosen  that  the  curve  repeats  itself  at  a  value  of  0  =  2?r, 
i.e.,  the  wave  is  periodic. 

Obviously,  if  the  distorted  wave  is  given  graphically  it  is 
always  possible  to  read  off  the  ordinate  corresponding  to  each 
abscissa  (Fig.  153). 


27T 


FIG.  153. 


The  problem  then  resolves  itself  into  finding  the  coefficients 
«o,  «i,  «n,  &o,  &i,  &»in  (103). 

To  do  this  a  mathematical  transformation  has  been  worked 
out  involving  convenient  integrations  and  the  fact  that  sines 
and  cosines  have  the  same  values  at  0  =  0  as  at  0  =  2?r  or  any 
multiple  of  27r,  that  is,  2?rn,  where  n  is  an  integer  number. 

To  find  a0  integrate  Eq.  (103)  between  0  and  2?r.     Thus, 


yds 


r^ 
I 

-  I 


a«de  + 


rzr 
I 


cos  BdB  + 


cos  nOde  -f  I     61  sin  6d0  + 


196 


1 


ri* 
I     an- 


bn  sin  n0d0. 


DISTORTED  WAVES  197 

From  what  has  been  said  above,  all  integrals  except  the  first 
must  be  zero.  Thus 

j      ydO  =  I      a0dB  =  a0(2ir  -  0)  =  27ra0. 

1    F* 
.'.  a0  =  -^~        yds. 

^ 
But    I     yds  is  the   area  of  the  curve  during  one  complete 

period  and  2ir  is  the  abscissa. 

.'.  a0  is  the  average  value  of  all  the  ordinates,  or  the  average 
value  of  y. 

To  determine  any  other  coefficient,  for  instance  a«,  Eq.  (103) 
is  multiplied  by  cos  nB  and  integration  is  again  carried  out  be- 
tween limits  0  and  2ir. 

In  this  case  it  is  also  remembered  that  the  integral  over  one 
period  of  any  product  of  sine  and  cosine  terms  is  zero. 

rr  r2*  r2* 

y  cos  nBdB  =  a0  I     cos  nBdB  +  «i  J     cos  nB  cos  BdB 

+ an  I     cos2  nBdB  +  fei  I     cos  nB  sin  BdB- 

+  fen  I      cos  nB  sin  n&dB. 

All  these  integrals  on  the  right-hand  side  must  be  zero  with 
the  exception  of 


cos2  nBdB, 

and  this  integral,  as  is  readily  seen,  is  =  TT. 

I  C2v 

.'.  an  =  -  I     y  cos  nBdB. 

But  J*y  cos  nBdB  is  the  area,  not  of  the  original  curve,  but  of 
another  curve  which  is  obtained  by  multiplying  each  value  of  y 
by  the  particular  value,  at  phase  angle  B}  of  cos  nB. 

Since  that  area  is  divided  by  TT  the  integral  must  be  just  twice 
the  average  of  the  instantaneous  values  of  ?/,  multiplied  by  cos  nB. 

.'.  an  =  2X  avg.  of  y  cos  nB  between  0  and  2ir. 
In  a  similar  way  all  values  of  fe  are  obtained  so  that, 
6n  =  2X  avg.  of  y  sin  nB  from  0  to  2ir. 

2» 

.'.  a0  =  avg.  (y)0 


198  ELECTRICAL  ENGINEERING 


ai  = 

2X 

avg. 

(y 

cos 

0) 

0 

a2  = 

2X 

avg. 

COS 

20)0' 

a3  = 

2X 

avg. 

(y 

COS 

30)1' 

an  = 

2X 

avg. 

(y 

COS 

I2' 
n0)!o 

2» 

61  = 

2X 

avg. 

(y 

sin 

?) 

0 

62  == 

2X 

avg. 

(y 

sin 

20 

)o 

bn  =  2X  avg.  (y  sin  n0)l0 

It  should  be  noted  that  dividing  the  curve  up,  say  every  10° 
from  0  to  360°,  37  readings  are  obtained.  It  is  better  then  to  use 
36  and  to  take  the  average  value  of  the  values  at  0°  and  360° 
instead  of  using  both  of  them. 

In  a  symmetrical  wave  only  those  harmonics  can  exist,  which, 
with  an  increase  of  the  angle  by  180°  or  TT,  reverse  the  sign  of  the 
function. 

This  is  only  the  case  when  n  is  an  odd  number.  Since,  if  n  is 
2,  4,  6,  etc.,  then  increasing  the  angle  by  TT  means  27r,  4?r,  6V,  etc., 
and  the  values  of  the  sine  and  cosine  are  the  same  for  a,  (a  +  2ir), 
(a  -f  4?r),  etc.,  whereas  if  n  =  1,  3,  5,  etc.,  we  get,  TT,  3r,  STT,  in 
which  the  sign  of  the  function  reverses. 

If  sin  a  is  positive,  then  sin  (a  +  TT)  is  negative. 

If  cos  a  is  positive,  cos  (a  +  v)  is  negative,  etc. 

Thus,  for  symmetrical  waves  such  as  are  given  by  alternators 
under  stable  conditions,  the  trigonometric  series  becomes : 

y  =  ai  cos  0  +  a3  cos  30  +  a5  cos  50  +.    .    .  -f-  61  sin  0 
+  63  sin  30  +  &5  sin  50  +     .... 

Obviously,  in  that  case,  it  suffices  to  analyze  one-half  a  wave 
only.1 

Problem  78.— Plot  the  wave, 

e  =  Ei  sin  6  +  E3  sin  (36  +  a), 
for 

El  =  1 
Ez  =  0.5 
a  =  30°, 

and  analyze  the  wave,  proving  that  the  analysis  gives  the  original  equation. 
Show  also  that  no  5th  harmonic  exists. 

1For  a  more  complete  discussion  of  this  method  of  wave  analysis  see 
STEINMETZ'S  "  Engineering  Mathematics." 


DISTORTED  WAVES 


199 


Tabulating : 


6° 

0 

10 

20 

30 

40 

50 

60 

70 

80 

90 

Ei  sin  0  

0.0 

0.174 

0  342 

0  50 

0  643 

0  766 

0  866 

0  94 

0  935 

1  0 

38  +  a  
Sin  (30  +  a)  

tfasin  (39  +  a)... 

30.0 
0.5 

0.25 
0  25 

60.0 
0.866 

0.433 
0  607 

90.0 
1.0 

0.5 
0  842 

120.0 
0.866 

0.433 
0  933 

150.0 
0.5 

0.25 
0  893 

180.0 
0.0 

0.0 
0  766 

210.0 
-0.5 

-0.25 
0  616 

240.0 
-0.866 

-0.433 
0  507 

270io 
-1.0 

-0.5 
0  485 

300.0 
-0.866 

-0.433 
0  567 

0° 

100.0 

110.0 

120.0 

130.0 

140.0 

150.0 

160.0 

170.0 

180.0 

Ei  sin  0  

0.985 

0.94 

0.866 

0.766 

0.643 

0.50 

0.342 

0.174 

0.0 

36  +  a  

330.0 

360.0 

30.0 

60.0 

90.0 

120.0 

150.0 

180.0 

210.0 

Sin  (30  +  a)  

-0.5 

0.0 

0.5 

0.866 

1.0 

0.866 

0.5 

0.0 

-0.5 

Et  sin  (36  +  a)..  . 

-0.25 

0.0 

0.25 

0.433 

0.5 

0.433 

0.25 

0.0 

-0.25 

e  

0.735 

0.94 

1.116 

1.2 

1.143 

0.933 

0.592 

0  174 

-0  25 

Analysis. — a0  must  be  zero  because  the  wave  is  symmetrical  above  and 
below  the  center  line.  The  coefficients  of  the  fundamental  cosine  and  sine 
waves  are  found  from 

ai  =  2  X  avg.  e  cos  0, 
bi  =  2  X  avg.  e  sin  6. 


e 

Cos  e 

e 

e  cos  0 

Sin  $ 

e  sin  0 

0 

1.0 

0.25 

0.25 

0.0 

0.0 

10 

0.985 

0.607 

0.598 

0.174 

0.1057 

20 

0.94 

0.842 

0.792 

0.342 

0.288 

30 

0.866 

0.933 

0.808 

0.5 

0.466 

40 

0.766 

0.893 

0.685 

0.643 

0.575 

50 

0.643 

0.766 

0.493 

0.766 

0.587 

60 

0.5 

0.616 

0.308 

0.866 

0.534 

70 

0.342 

0.507 

0.173 

0.94 

0.477 

80 

0.174 

0.485 

0  .  0844 

0.985 

0.478 

90 

0.0 

0.5fc7 

0.0 

1.0 

0.567 

100 

-0.174 

0.735 

-0.128 

0.985 

0.725 

110 

-0.342 

0.94 

-0.322 

0.94 

0.885 

120 

-0.5 

1.116 

-0.558 

0.866 

0.966 

130 

-0.643 

1.2 

-0.772 

0.766 

0.920 

140 

-0.766 

1.143 

-0.875 

0.643 

0.735 

150 

-0.866 

0.933 

-0.808 

0.5 

0.466 

160 

-0.94 

0.592 

-0.556 

0.342 

0.202 

170 

-0.985 

0.174 

-0.171 

0.174 

0.0303 

180 

-1.0 

-0.25 

-0.25 

0.0 

0.0 

200 


ELECTRICAL  ENGINEERING 


The  sum  of  the  18  cosine  readings,  using  the  average  of  0°  and  180°  as 
one,  is  -  0.2486  and  the  average  value  is  -  0.0138. 
Thus, 

ai  =  2  X  avg.  =  -  0.0276. 

Similarly,  the  sum  of  the  sine  readings  is:  9.007 

The  average  is  0.5004, 

Thus, 

61  =  2  X  avg.  =  1.0008. 

The  coefficients  of  the  3d  harmonics  are  found  from, 
a3  =  2  X  avg.  e  cos  30, 
63  =  2  X  avg.  e  sin  36. 


e° 

30 

Cos  30 

e  cos  30 

Sin  30 

e  sin  30 

0 

0 

1.0 

0.250 

0.0 

0.0 

10 

30 

0.866 

0.525 

0.5 

0.304 

20 

60 

0.5 

0.421 

0.866 

0.730 

30 

90 

0.0 

0.0 

1.0 

0.933 

40 

120 

-0.5 

-0.447 

0.866 

0.774 

50 

150 

-0.866 

-0.664 

0.5 

0.383 

60 

180 

-1.0 

-0.616 

0.0 

0.0 

70 

210 

-0.866 

-0.439 

-0.5 

-0.254 

80 

240 

-0.5 

•     -0.242 

-0.866 

-0.420 

90 

270 

0.0 

0.0 

-0.1 

-0.567 

100 

300 

0.5 

0.368 

-0.866 

-0.637 

110 

330 

0.866 

0.815 

-0.5 

-0.470 

120 

360 

1.0 

1.116 

0.0 

0.0 

130 

390 

0.866 

1.040 

0.5 

0.600 

140 

420 

0.5 

0.571 

0.866 

0.990 

150 

450 

0.0 

0.0 

1.0 

0.933 

160 

480 

-0.5 

-0.296 

0.866 

0.513 

170 

510 

-0.866 

-0.151 

0.5 

0.087 

180 

540 

-1.0 

0.250 

0.0 

0.0 

The  sum  of  the  18  cosine  readings  is  2.251.      The  average  is  0.125. 

.'.  a3  =  2  X  avg.  =  0.25. 

The  sum  of  the  sine  readings  is  3.899.     The  average  is  0.2165; 
.*.  63  =  0.433. 

The  exercise  of  proving  that  no  5th  harmonic  exists  is  left  for  the  student. 
Summing  up  the  values  already  obtained,  the  equation  may  be  written : 

e  =  -  0.0276  cos  6  +  0.25  cos  30  +  1.0008  sin  0  -  0.433  sin  30 
which  is,  approximately, 

y  =  sin  0  +  0.433  sin  30  +  0.25  cos  30. 


DISTORTED  WAVES 


201 


The  second  and  third  terms  may  be  combined  or  added,  being  in  quad- 
rature, by  the  vectorial  method  in  which 


where 


Thus, 


A  sin  6  +  B  cos  0  =  \/A2  +  B2  sin  (0  +  a), 


tan  a  =  —A 
A. 


0.433  sin  30  +  0.25  cos  30  =  Vo.188  +  0.0625  sin  (30  +  a) 

=  0.5  sin (30  +  a),  where  a  =  tan"1  Q-TOQ  =  30°. 
The  complete  wave  is,  therefore, 

e  =  sin  0  -{-  0.5  sin  (30  +  30°). 

The  wave  is  shown  plotted  in  Fig.  154,  in  which  also  the  component 
waves  are  indicated  by  the  dotted  lines. 


FIG.  154. 


CHAPTER  XXX 


MECHANICAL  STRESSES  IN  TRANSFORMERS 

It  is  recollected  that  a  mechanical  force  is  exerted  on  a  con- 
ductor carrying  current  if  it  is  placed  properly  in  a  magnetic  field, 
the  force  being  1  dyne  per  cm.  of  conductor  per  abamp.  in  a  field 
intensity  of  1  line  per  sq.  cm.  provided  the  field  is  at  right  angles 
to  the  conductor. 

Referring  to  Fig.  155,  which  represents  the  cross-section  of  a 
transformer,  it  is  evident  that  the  main  flux  which  interlinks  with 
both" the  primary  and  the  secondary  windings  and  is  confined  to 
the  iron  does  not  cut  through  any  part  of  the  windings  carrying 
current,  but  that  the  leakage  flux  more  or  less  completely  cuts  the 
windings  and  therefore  is  responsible  for  a  force  which  tends  to 
warp  the  coils  out  of  shape  and  thus  to  damage  them.  The  deter- 
mination of  the  mechanical  stresses  resolves  itself  therefore  largely 

into  the  calculation  of  the  leakage 

core  flux  or  leakage  inductance  of  the 

transformer. 

To  calculate  the  leakage  induc- 
tance of  the  secondary  coil,   con- 
•  sider  this  made  up  of  the  interlink- 
.  ages  of  flux  with  turns  in  the  space 
FIG.  155.  occupied    by    the    secondary    coil 

itself,  plus  the  interlinkages  of  the 

flux  between  the  coils  with  all  of  the  secondary  turns.     Similarly 
with  the  primary. 
Approximation  of  the  Leakage  Inductance  of  the  Secondary.— 


-  turns,  where 


In  Fig.  155,  a  portion  of  the  coil  of  depth  x,  has 

a  is  the  total  depth  of  the  coil,  and  N2  is  the  total  number  of 
secondary  turns  on  1  leg  of  the  transformer. 

The  magnetomotive  force  of  this  part  of  the  coil  is 

t  r   *r    x 

m.m.fx  =  I2N2  — 
a 

The  flux  which  this  m.m.f.  produces  is 
flux  = 


202 


MECHANICAL  STRESSES  IN  TRANSFORMERS    203 

where  p  is  the  reluctance,  and  p  =  —  —  =  —  §-    when    we    con- 

area       max 

sider  only  the  flux  which  passes  through  the  small  area  of  width 
dx  and  length  m.  m  is  the  length  of  a  turn  at  distance  x  in  Fig. 
155.  It  is  almost  impossible  to  determine  accurately  the  length 
Z0.  It  is  the  equivalent  length  of  the  lines  of  force  which  going 
through  section  mdx  return  upon  themselves.  Part  of  these  lines 
can  be  readily  traced.  They  go  almost  straight  across  the  trans- 
former windings  of  length  h;  then  they  spread  apart,  and  the 
equivalent  length,  as  a  result,  is  relatively  short.  Then,  the 
majority  of  the  lines  enter  the  iron  and  their  reluctance  is  insig- 
nificant. Some,  however,  enclose  the  winding  that  is  outside  of 
the  iron  and  these  meet  with  considerable  reluctance.  Therefore, 
it  might  be  fairly  conservative  to  assume  Z0,  the  equivalent  length, 
as  I  the  height  of  the  "window"  of  the  transformer. 

If  mz  is  the  mean  length  of  a  secondary  turn,  this  may  be  sub- 
stituted for  m,  thus 

I 
p  ' 


Then  the  flux  in  any  elementary  band,  dx,  is 

m2dx  x 


d<f>' 


This  flux  interlinks  with  —  Nz  turns.     Therefore,  the  interlink- 


/>»        /y» 

ages  with  the  flux  =  ^irl^Nz  "**"""     -  —  N%,  and  the  inductance  due 
to  the  interlinkages  within  the  space  occupied  by  the  coil  is 

(104) 


4 


To  determine  the  inductance  due  to  the  flux  in  the 
gap  between  coils,  consider  Fig.  156  which  shows  a 
section  through  one  side  of  the  coils.  The  current  is 
oppositely  directed  in  the  two  coils,  as  indicated  by 
dots  and  crosses.  On  a  1:1  basis,  the  turns  and 
currents  in  the  two  coils  are  equal,  and  the  figure  may 
be  regarded  as  merely  showing  a  section  through  a 
single  coil,  of  Nz  turns,  or  of  Ar2/2*amp.-turns. 

The  area  of  the  core  of  this  imaginary  coil  will  be  6m3,  where 
ra3  is  the  mean  circumference  between  the  actual  coils,  and  6  is 


204 


ELECTRICAL  ENGINEERING 


the  distance  between  them.     The  flux  produced  in  this  region  by 
the  m.m.f.,  /2N2,  is  then 


o  X  4-n  Ar22  iw3,  due  to 


The  number  of  interlinkages  is 

This  represents  an  inductance  of  L"2 

the  secondary  coil,  since  half  of  the  inductance  is  due  to  the  pri- 
mary and  the  other  half  due  to  the  secondary. 

The  total  secondary  inductance  of  this  coil  is  then 


L2  = 


Lff 
2    = 


and  the  primary  inductance  is,  similarly, 

T         27rAVr9       c 
Li-    —  j  —  [2rai^ 

where  c  is  the  depth  of  the  primary  coil. 


Since  ^  = 


N 


former,  referred  to  the  primary  is 

L  =  ^y1-^!^  + 
If  two  legs  are  in  series,  L  (tote/) 
or,  if  in  parallel,  L  (tota0  =  -• 
In  practical  units, 

L  =  32  X  10-9  ~ 


2>  the  total  inductance  on  1  leg  of  the  trans- 

n.  (105) 


2L, 


m2 


henrys    '  (106) 


where  the  dimensions  are  in  inches. 

The  same  reasoning  may  be  applied 
to  a  core-type  transformer  in  which 
the  coils  are  differently  arranged,  for 
example,  as  in  Fig.  157. 

Here  are  two  secondary  coils,  with 
the  primary  placed  between  them. 
Consider  the  primary  as  if  made  up 
of  two  equal  coils,  separated  by  a 
dividing  line  shown  dotted.  The 
calculation  should  then  be  made  of  the  combined  inductance  of 
the  secondary,  S',  and  one-half  of  the  primary,  which  are  grouped 


MECHANICAL  STRESSES  IN  TRANSFORMERS    205 

together  as  A  in  the  figure,  and  similarly,  the  secondary  S"  and 
the  other  half  of  the  primary  grouped  as  B. 
From  (105),  the  inductance  for  A  is, 


U 


and  for  B, 


in  which 


]2|~,w,/ 

i  m 


w'i  =  mean  length  of  inside  one-half  primary  turn 
w"i  =  mean  length  of  outside  one-half  primary  turn 

m'2  =  mean  length  of  inside  secondary  turn 
m"z  =  mean  length  of  outside  secondary  turn 

m3  =  mean  length  of  inside  gap 

ra4  =  mean  length  of  outside  gap 
w"i  =  2rai 


m3  +  w4  =  2m. 
If  coils  are  symmetrical,  mi 


m2. 


Supplying  all  of  these  values,  the  total  inductance  is 

L  =  L'+L"  =  -^r-\mi~  +  m2  %  +  ra&lcm., 
I     L      6  3  J 

where  Ni  is  the  number  of  turns  in  half  the  primary  coil.     If  TI 
is  the  number  of  primary  turns  per  leg  of  the  core, 


c  a 

-  +  m2 


If  dimensions  are  in  inches, 
T        16  TYr      c 

•Lj     -,  /-vn  T 


cm.  per  leg. 
mb] 


FIG.  158. 


In  a  similar  manner,  shell-type  transformers  may  be  dealt  with. 
Such  a  transformer  is  shown  in  Fig.  158.  In  this,  let  m  =  mean 
length  of  1  turn,  NI  =  number  turns  in  half  of  a  primary 


206  ELECTRICAL  ENGINEERING 

coil,  =  one-quarter  total  primary  turns.  Using  the  same 
reasoning  as  with  core-type  transformers,  the  inductance  of  a 
unit  combination,  A,  in  the  figure,  is 


—  [|  +  I  +  b]  cm.  (107) 


Note  that  m  ^  +  «•  +  Z     is  the  equivalent  area,  whence  the 
total  inductance  is 


SL  =          j        |g  +  F+&|  cm. 


In  inch  units, 

128 


Calculation  of  Stresses.  —  Under  ordinary  conditions  of  load, 
these  would  not  be  excessive,  but  for  maximum  current,  as  in  the 
case  of  short-circuit,  or  heavy  transient  currents  from  switching, 
they  may  be  very  great.  Calculation  may  properly  be  based  on 
the  short-circuit  current,  remembering  again  that  a  wire  1  cm. 
long,  carrying  10  amp.  (unit  current),  if  placed  perpendicular 
to  a  field  of  1  line  per  sq.  cm.,  is  repelled  by  a  force  of  1  dyne; 
or  the  force  in  dynes  =  BI'l,  where  I'  is  expressed  in  absolute 
values  —  abamperes. 

If  the  flux  density  in  the  gap,  &,  between  coils,  is  Bmax  then  it 
may  be  assumed  that  the  average  density  of  the  flux  leaking 

D 

through  the  coils  themselves  is  —  ^,  which  is  then  the  average 

density  of  the  flux  passing  through  the  coils  of  any  section  A, 
Fig.  158,  and  the  force  per  turn  on  any  coil,  will  be 

Ft  =  -^  X  1\  X  m  dynes 

B 


max 


98f  grams' 

where  m  is  the  mean  length  of  the  turn. 
If  /2  is  in  amperes, 


™  Bmax        z 

Ft  =  ——  grams  per  turn. 


Let  the  effective  value  of  the  short-circuit  current  be  72,  and 
let  the  total  secondary  turns  be  T2,  then  the  turns  in  a  half  coil 

(Fig.  158)  are      - 


MECHANICAL  STRESSES  IN  TRANSFORMERS    207 
The  maximum  value  of  the  force  will  be 


2  X  9810      v   4 

=     8X9810m  grams  (108> 

or  in  the  case  of  the  primary  short-circuit  current  / 

m  V2ITmBm 
8  X  9810 

where  /  is  the  effective  value  of  the  primary  short-circuit  current 
and  T  the  total  number  of  primary  turns. 

The  leakage  flux  must,  in  the  case  of  short-circuit,  be  the  main 
flux  (neglecting  the  flux  due  to  the  voltage  which  is  consumed 
by  the  ohmic  drop),  if  it  is  assumed  that  the  generating  station 
is  large  and  the  voltage  impressed  upon  the  transformer  is 
normal  even  though  the  transformer  is  short-circuited.  (See  note.) 

The  maximum  value  of  the  flux  between  a  group  of  coils  is 
obtained  by  multiplying  the  maximum  value  of  the  flux  density 
Bm  by  the  equivalent  area  as  given  in  (107). 

That  is 


The  group  contains  in  this  case  one-quarter  of  the  turns  and 

ET 

the  voltage  per  group  is  -r  where  E  is  the  effective  value  of  the 

impressed  e.m.f. 

The  relation  between  the  maximum  value  of  the  flux  and  the 
voltage  is  given  by  the  well-known  relation 


Substituting  this  in  (109) 

(110) 


The  average  value  of  the  force  is  obviously  one-half  of  the 
maximum  value. 


208  ELECTRICAL  ENGINEERING 

The  force  between  the  coils  is  proportional  to  the  rating 
assuming  the  same  regulation. 

NOTE. — The  actual  flux  enclosed  by  the  secondary  turns  depends  upon 
the  terminal  voltage  and  the  ir  drop. 

At  short-circuit  the  secondary  terminal  voltage  obviously  is  zero.  Thus 
if  as  a  limiting  case  the  ir  drop  is  neglected  the  secondary  winding  encloses 
no  flux. 

As  long  as  it  is  assumed  that  the  primary  voltage  is  normal  voltage  and 
that  the  ir  drop  is  again  neglected  the  primary  coil  encloses  the  same  flux 
during  the  short-circuit  as  it  does  at  no-load.  The  path  of  the  flux  must 
therefore  be  essentially  different.  In  the  latter  case  it  traversed  the  two 
windings  and  is  therefore  mainly  in  the  iron,  while  in  the  former  case  it 
must  traverse  only  one  winding — the  primary.  Thus  the  flux  must  find 
its  way  between  the  primary  and  secondary  coils  and  is  thus  the  so-called 
leakage  flux. 


CHAPTER  XXXI 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN 

Type. — Transformers  may  be  classified  as  belonging  either  to 
the  "core  type"  or  the  " shell  type." 

Core-type  transformers  frequently  have  a  single  magnetic 
circuit  of  rectangular  form.  On  the  two  vertical  sides  of  this 
core  are  placed  the  windings,  each  side  being  provided  with 
half  of  the  primary  and  half  of  the  secondary  coils,  the  low- 
voltage  coils  usually  being  placed  next  to  the  core  (Fig.  159). 

Shell-type  transformers  usually  have  a  multiple  magnetic 
circuit  the  coils  being  placed  upon  a  central  core,  the  outer  limbs 
of  which  extend  around  the  coils,  somewhat  resembling  a  shell 
(Fig.  160).  As  illustrated  diagrammatically  in  the  figures,  it  is 


Secondary 


Core 


Primary 


FIG.  159. 


FIG.  160. 


seen  that  the  coils  of  the  core-type  transformer  have  the  form 
of  a  cylindrical  shell,  while  those  of  the  shell  type  are  in  the  form 
of  discs.  The  former  lend  themselves  readily  to  designs  of  great 
mechanical  strength,  while  the  latter  tend  to  be  mechanically 
weak. 

The  present  tendency  seems  to  be  more  and  more  toward 
the  core  type,  and  it  remains  for  the  superiority  of  the  shell 
type  to  be  demonstrated  in  any  given  case  in  order  to  justify  its 
existence  at  all. 

Recently  transformers  having  a  multiple  magnetic  circuit  have 
been  introduced.  The  coils  are  of  the  cylindrical  form  placed 
around  the  central  core.  Thus,  this  is  called  the  cruciform  type. 
14  209 


210 


ELECTRICAL  ENGINEERING 


An  important  consideration  with  respect  to  the  choice  of  type 
is  the  method  of  cooling  the  transformer.  Core-type  transform- 
ers are  usually  immersed  in  oil  in  such  a  way  as  to  provide  free 
circulation  of  the  oil  about  all  surfaces  of  the  coils  and  core. 
The  oil  then  receives  the  heat  and  carries  it  to  the  outside  case 
which  is  frequently  corrugated  to  present  greater  effective 
surface  to  the  outer  air. 

Shell-type  transformers  are  cooled  by  the  above  method,  but 
more  frequently  this  is  augmented  by  the  addition  of  coils  of 
pipe  through  which  is  forced  a  stream  of  cool  water.  These 
coils  are  placed  in  the  oil  above  the  transformer. 

The  addition  of  the  cooling  water  is  essentially  a  feature  of 
large  transformers,  since  they  have  less  area  of  possible  cooling 
surface  per  unit  volume  than  have  smaller  units. 

A  common  form  of  the  shell  type  is  known  as  the  air-blast 
type.  The  method  of  cooling  consists  in  forcing  a  continuous 
blast  of  cool  air  up  through  the  ducts  with  which  the  core  is 
provided,  and  between  and  around  the  coils. 

Efficiency. — Transformers  are  not  designed  to  give  the  highest 
possible  efficiency  as  this  would  involve  too  great  an  expense  in 
materials  and  manufacture,  but,  rather,  the  highest  practical 
efficiency,  so  as  to  meet  competition  both  in  price  and  in  quality. 

Consequently,  from  results  obtained  in  practice,  it  is  easy  to 
construct  a  table  of  efficiencies  which  might  reasonably  be 
expected  of  various  sizes  of  transformers  of  moderate  voltages, 
say  up  to  10,000  volts.  This  table  is  as  follows : 


Efficiency 

25  cycles 

60  cycles 

1 

94.0 

96.0 

5 

96.5 

97.5 

10 

97.0 

98.0 

50 

98.0 

98.5 

200 

98.0 

98.5 

Knowing  the  approximate  efficiency  of  the  transformer  which 
is  to  be  designed,  the  total  losses  are  of  course  also  known. 
For  example,  let  it  be  required  to  design  a  10-kw.,  60-cycle, 
200%oo-v°lt  core-type  lighting  transformer.  The  efficiency  is 
to  be  about  98  per  cent.  The  losses  are  2  per  cent.,  or  0.02  X 
10,000  =  200  watts. 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  211 

Losses. — These  losses  are  made  up  of  the  I2r  loss  in  the  copper 
windings  and  the  hysteresis  and  eddy  current  losses  in  the  iron 
core  and  windings. 

Maximum  efficiency  is  obtained  at  that  load  for  which  the 
copper  and  iron  losses  are  equal.  It  becomes  a  matter  of  choice 
in  design  as  to  what  ratio  shall  be  given  these  losses  or  at  what 
load  they  shall  be  equal.  Thus,  for  power  purposes,  the  copper 
and  iron  losses  should  be  about  equal  at  full-load,  giving  maxi- 
mum efficiency  at  full-load.  For  lighting  purposes,  however, 
owing  to  the  peculiar  conditions  of  operation,  this  is  not  generally 
desirable.  A  lighting  transformer  carries  full-load  only  for  a 
very  small  period  during  each  24  hr.,  while  the  rest  of  the  time  it 
is  operating  practically  at  no-load.  Thus  the  copper  loss  is  quite 
small  even  with  a  large  value  of  72r,  while  the  core  loss  is  larger 
since  it  is  continuous  through  the  whole  day.  It  would  be  better, 
therefore,  to  make  the  copper  loss  relatively  greater  than  the 
core  loss,  at  full-load,  and  thus  reduce  the  total  losses  for  the 
daily  operation.  Fairly  good  values  to  choose  for  these  losses 
are:  copper  loss  =  60  per  cent.,  core  loss  =  40  per  cent,  of  the 
total  loss. 

In  the  example  considered, 

copper  loss  =  Pr  =  200  X  0.60  =  120  watts, 
core  loss  =  200  X  0.40  =  80  watts. 

The  core  loss  may  be  further  divided  between  loss  due  to 
hysteresis  and  loss  due  to  eddy  currents.  The  former  is  usually 
larger  because  it  depends  on  the  magnetic  quality  of  the  iron  or 
steel  used,  whereas  the  latter  depends  largely  on  the  degree  of 
thinness  of  the  laminations  of  the  core,  and  this  may  be  carried 
to  any  extent  mechanically  practical.  Values  of  hysteresis  and 
eddy  current  losses  when  silicon  steel  laminations  .014  in.  thick 
are  used  are: 

hysteresis  loss  =  0.7  watt  per  Ib.  at  60  cycles, 
eddy  current  loss  =  0.3  watt  per  Ib.  at  60  cycles, 

when  the  maximum  induction  density  is  64,500  lines  per  sq.  in. 
(10,000  lines  per  sq.  cm.). 

Since  1  cu.  in.  of  this  material  weighs  0.28  Ib.,  the  loss  per  cu. 
in.  at  60  cycles  and  64,500  lines  per  sq.  in.  is: 

hysteresis  loss  per  cu.  in.  =  0.28  X  0.7  =  0.196  watt, 
eddy  current  loss  per  cu.  in.  =  0.28  X  0.3  =  0.084  watt, 
total  core  loss  per  cu.  in.  =  0.28  watt. 


212 


ELECTRICAL  ENGINEERING 


Hysteresis  loss  for  any  frequency  and  density  is  given  ap- 
proximately by  the  equation, 


hyst.  loss  =  Wh  =  0.196  X  ^  X 

where  V  =  volume  of  iron. 
Similarly,  eddy  current  loss  is 

/ 


V, 


0.084  X 


From  these  two  equations  and  the  core  loss  which  is  given, 
the  volume  may  be  obtained  for  any  value  of  B.  Assuming,  as 
will  later  be  done,  that  B  =  70,000,  in  the  example,  * 


80  ^  [0.196  X  (1.086)1-6  +  0.084(1. 086) 2]  = 

80 
0.2205  +  0.099 


=  250  cu.  in. 


And  the  hysteresis  loss  is  Wh  =  0.196  X  1.125  X  250  =  55.2 
watts,  and  the  eddy  current  loss  is  We  =  0.099  X  250  =  24.8 
watts. 


0  5  10  15  20 

Volume  per  Watt  Hysteresis  Loss  Cu,  In. 
0              0.1             0.2             0.3            0.4  0.5  0.6 

Watts  per  Cubic  Inch 

FIG.  161. 

B  and  V. — The  relation  between  B  and  V  is  shown  by  the 
following  curve,  Fig.  161,  from  which  it  is  evident  that  values 
of  B  should  lie  between  50,000  and  90,000. 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  213 
Hysteresis  and  Eddy  Current  Loss  per  Cubic  Inch.— 


(!)/  =  60 


B 

50,000 

60,000 

70,000 

80,000 

90,000 

100000 

B 

0  775 

0  93 

1  085 

1  24 

1  395 

i    fit) 

64,500  
/     B     \i.e 

Of\K 

OQQ 

1     19^ 

1    d.9 

Iijt* 

2f\f 

\64,500/ 
TF        OIQfi    (    B     V6 

0197 

01  79  f\ 

099O 

097ft 

.  to 
00.^0 

.  uo 

OAfyy 

Wk  -  0.19G   ^4 
1    B     y 

OAfll 

OftAC 

11  ft 

.^/o 

1£A 

.o^o 

IQr 

.4UZ 
241 

\64,500/ 
TF       n  ns4  /    B    \  2 

.OU1 

OfkKfJK 

.oDO 
OO797 

.  lo 

Onoo 

.  O4 
01  9QJ. 

.»O 
01  AQQ 

.41 
Oonne 

W.  -  0.084   ^64j50Qj 
TF 

7  • 

0.1775 

0.2452 

0.319 

0.4074 

.  iDoy 
0.5069 

0.6045 

(2) /  =  25;        =  0.417;  =  0.174 


WK  

0.053 

0.072 

0.0917 

0.116 

0.143 

0.1675 

W,  
W 

0.0088 
0.0618 

0.0127 
0.0847 

0.0172 
0.1089 

0.0225 
0.1385 

0.0285 
0.1715 

0.0384 
0.2059 

v  

As  a  matter  of  fact  the  usual  limits  are: 

for  60  cycles,  B  lies  between  60,000  and  75,000, 
for  25  cycles,  B  lies  between  80,000  and  90,000. 
In  the  example,  let  B  =  70,000,  which  will  be  taken  as  a  trial 
value. 

From  Fig.  162  the  volume  per  watt  loss  by  hysteresis  is  4.55 
cu.  in.  The  total  volume  of  iron  is  4.55  X  55.2  watts  =  250  cu.  in. 
Magnetizing  Current. — Having  chosen  a  suitable  value  of  B, 
we  can  at  once  find  out  the  required  number  of  ampere-turns  per 
inch  length  of  magnetic  circuit,  from  the  saturation  curve,  p. 
Let 

M o  =  ampere-turns  per  inch  and 
I  =  length  of  magnetic  circuit. 

Then  total  ampere-turns  =  MQl  =  -\/2imt,  where  -\/2im  =• 
maximum  value  of  magnetizing  current  and  t  =  number  of  turns 
on  the  primary. 

Using  the  fundamental  equation  for  e.m.f., 

E  =  4.44  ft*  X  10-8  =  4A4ftBA  X  10~8, 


214  ELECTRICAL  ENGINEERING 

the  magnetizing  volt-amperes  are 

1-1  " 


X  108 

=  vfBAMol  X  10~8  =  irfBMoV  X  10~8, 
since 

4.44  =  \/2ir,  and  V=Al. 

The  percentage  magnetizing  current  is  obtained  by  dividing  by 
El,  thus, 

T  =10uXkw.' 

In  the  example,  M 0  is  found  to  be  6.5.  Therefore,  substituting 
known  values  into  the  equation, 

im       3.14  X  60  X  70,000  X  6.5  X  250 

T :  io"  x  10  l5' 

or  approximately  2  per  cent. 

This  is  a  reasonable  value.  In  practice,  magnetizing  currents 
range  from  2  to  8  per  cent.,  being  larger  in  smaller  transformers 
and  at  lower  frequencies. 

Number  of  Turns,  Total  Flux,  Area,  and  Length  of  Magnetic 
Circuit. — Returning  to  the  fundamental  e.m.f.  equation,  it  is  seen 
that  turns  and  flux  are  both  unknown.  A  practical  limit  in  help- 
ing to  decide  what  value  to  assign  to  either  one  of  these  unknowns 
is  found  from  the  fact  that  the  number  of  turns  should  depend 
upon  the  voltage.  While  it  would  not  be  safe  to  allow  too  great 
a  difference  of  potential  to  exist  between  adjacent  turns,  this 
consideration  is  not  the  deciding  feature.  The  choice  of  number 
of  turns  is  governed  largely  by  cost  considerations.  From  prac- 
tice it  is  known  that  volts  per  turn  should  lie  between  0.4  X  \/kw. 
and  0.6  X  A/kw.  in  core-type  transformers.  The  former  value 
is  more  suitable  for  distribution  transformers  when  it  is  desirable 
to  keep  down  the  core  loss,  while  the  latter  is  suitable  for  power 
transformers.  The  value  for  shell  type  is  from  two  to  three 
times  as  great. 

In  the  example,  it  will  be  assumed  that  volts  per  turn  =  0.5  X 
=  1.56. 

Then,  turns  on  primary 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  215 
and  flux 

-  586>000' 


area 

$       586,000 

=  A   =B=:  "TpOO"  ==  8'37  sq' 
and  length, 


Resistance,  Length  of  Mean  Turn,  Total  Length  and  Size  of 
Windings.  —  Returning  now  to  the  windings,  it  is  possible  at  first 
to  calculate  the  primary  resistance,  since  the  copper  loss  and  the 
current  are  known. 

In  the  example  Pr  =120  watts.  This  must  be  divided  be- 
tween primary  and  secondary,  and  half  may  be  assigned  to  each, 
as  a  reasonable  approximation. 

Thus  primary 


Also, 


Pr  =          =  60  watts. 

W       10,000 
7l  =  Yi  ="  "2000T  =  5 

60 

.'.  #1  =  ^  =  2.4  ohms. 
&o 

Knowing  the  resistance  and  number  of  turns,  the  size  of  wire 
may  be  found  when  the  mean  length  of  one  turn  is  estimated. 
As  a  basis  for  this,  the  cross-sectional  area,  A,  of  the  core  is  known, 
and  experience  tells  about  how  much  space  is  necessary  for  insu- 
lation between  core  and  coils  and  for  circulation  of  the  cooling 
oil  between  the  coils.  Also,  since  the  heat  generated  in  the  in- 
terior of  the  coils  has  to  pass  through  the  thickness  of  copper  and 
insulation,  it  will  be  unwise  to  make  the  coils  too  thick. 

Practical  thickness  of  insulation  against  voltage  is  given  in 
the  following  table. 

TABLE  VII 

,.  ,.  Insulation 

Volta  thickness  (mils) 

110  40 

440  50 

1,000  70 

2,300  100 

6,600  180 

16,000  260 


216 


ELECTRICAL  ENGINEERING 


For  circulation  of  oil,  space  of  not  less  than  J£  m«  width  should 
be  allowed.  This  width  is  governed  by  the  height  of  the  coils. 

Thickness  of  the  coils  should  hardly  exceed  1  in.,  but  may 
reasonably  be  %  in. 

Applying  this  procedure  to  the  example,  it  is  found  that  with 
an  area,  A  =  8.37  sq.  in.  of  iron,  the  gross  area  occupied  by  the 
laminations  will  be  about 


If  this  area  is  in  the  form  of  a  square,  the  side  of  the  square 

will  be  -\/9l3  =  3.05  in.     Fig.  162  is 

next  drawn,  showing  the  relative 
positions  of  coils,  core,  insulation,  etc. 
In  this  case,  the  length  of  mean  turn 
of  the  secondary  winding  is 

L2  =  4  X  3.05  +  2^(0.25  + 
0.04  +  0.375)  =  16.4  in. 

Since  the  secondary  winding  is 
nearest  the  core,  its  features  will  be 
discussed  first,  thus  avoiding  any 
error  in  the  final  determination  of 


•TI 

,~\ 

t75- 

^ 

V 

-f 
-1 

1 

25, 
«=  3.05  —  :=> 

Core 

^  —       — 
1 

Primary 

FIG.  162. 


the  mean  length  of  primary  turn. 
Total  length  of  secondary  is 


16  4 
X  <2  =  -TH-  X 


12 


ft. 


In  general, 


100 
2000 


X  1280  =  64. 


In  practice,  however,  it  is  found  convenient  to  put  the  two 
primary  coils  in  series  and  the  two  secondary  coils  in  parallel  to 
obtain  the  20: 1  ratio. 

If  this  procedure  is  adopted,  the  voltage  impressed  on  one 

,    2000 
primary  coil  is  -y-  =  1000,  while  the  whole  secondary  voltage 

of  100  will  be  across  each  of  the  secondary  coils. 
The  secondary  turns  per  coil  will  then  be 


100 
1000 


X  640  =  64 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  217 

and  each  coil  will  carry  half  of  the  total  secondary  current    or 

10,000 

=  50  amp. 


Total  length  of  each  secondary  coil  is  then 

L2  X  t,  =  ~  X  64  =  87.3  ft. 

27? 

Resistance  per  1000  ft.  of  secondary  coil  =     AQi0* 

U.Uo7o 

Since  the  coils  are  connected  in  parallel,  the  resistance  of  one 
coil  is  2RZ. 

The  resistance  of  a  secondary  coil  is  obtained  from  the  fact 

60 
that  the  secondary  copper  loss  per  coil  is  -^  =  30  watts. 

Thus,  the  resistance  of  each  secondary  coil  is 

30 

=  °-012  onm- 


Resistance  per  1000  ft.  of  conductor  is  n  '    7Q  =  0.1375  ohm. 

U.Uo/o 

This  corresponds  to  an  area  of  0.07  sq.  in. 

The  conductor  chosen  must  be  of  copper  strip,  of  rectangular 
cross-section.  In  using  strip,  the  practical  dimensional  limits 
are  about  0.1  in.  in  thickness  and  0.5  in.  in  width.  These 
dimensions  give  an  area  of  0.05  sq.  in.  If  greater  area  is  re- 
quired, any  number  of  strips  may  be  wound  in  parallel.  Each 
strip  is,  however,  insulated,  usually  with  double  cotton  covering, 
to  prevent  too  great  eddy  current  loss  in  the  copper. 

In  the  present  case  there  will  be  two  strips  required,  each  of 
0.1  X  0.35-in.  section. 

With  insulation,  the  dimensions  of  the  double  conductor  become 
0.36  X  0.22  in. 

It  will  be  seen  that  the  most  practical  arrangement  of  the 
turns  will  be  to  have  two  layers  deep  and  32  turns  per  layer. 
Then  the  thickness  of  the  coil  becomes  2  X  0.22  in.  =  0.44  in.; 
the  length  of  the  coil  is  32  X  0.36  in.  X  11.5  in. 

The  corrected  mean  length  of  turn  is 

L2  =  4  X  3.05  +  27r  (0.25  +  0.04  +  0.22)  =  15.4  in. 

I  K  A    y   f\A 

Total  length  =  -  ^~     -  82  ft. 


Resistance  per  1000  ft.  =     ^  =  0.147  ohm. 


0  1375 
Corrected  cross-section  is  X  0.07  sq.  in.  =  0.0655  sq.  in. 


218  ELECTRICAL  ENGINEERING 

Maintaining  the  same  thickness,  i.e.,  0.1  in.  the  width  of  the 
strip,  with  insulation,  now  becomes  0.332  in.,  and  the  coil  length 
is  32  X  0.332  =  10.6  in. 

The  mean  length  of  the  primary  turn  may  now  be  found.     It  is 

Ll  =  4  X  3.05  +  2T  (0.25  +  0.04  +  0.44  +  0.04  + 

0.25  +  0.1  +  0.375)  =  12.2  +  2w  X  1.495  =  21.6  in. 

Total  length  of  primary  is  then 


/ 

21.6  X  ||  =  21.6  X  =  2304  ft. 

r>   .  24- 

Resistance  per  1000  ft.  of  primary  is  2304  =  2394.  = 

ohms. 

Referring  to  wire  tables,  this  resistance  is  found  to  be  nearly 
that  of  No.  10  B.  &  S.,  which  has  resistance  of  1.18  ohms  per  1000 
ft.  at  65°C. 

If  now  it  should  be  desirable  to  use  copper  strip  for  the 
primary  winding,  the  requisite  area  may  be  found  by  comparison 
with  that  of  No.  10  wire.  Thus, 

1  18 
area  =    -        X  0.00815  =  0.00922  sq.  in. 


In  this  case,  however,  it  will  be  practical  to  use  No.  10  wire. 
Wire  larger  than  No.  10  is  not  generally  used,  but  smaller  sizes 
are  preferable  to  rectangular  strip. 

Choosing  then  No.  10  wire  the  space  which  the  640  turns  of 
each  coil  will  occupy  must  be  determined. 

With  a  layer  10.3  in.  long  there  will  be 

10  3 
Q  IQ2  =  100  turns  per  layer, 

and 

640 

100  =:  6'4  layers' 

Obviously,  the  best  arrangement  of  these  640  turns  will  be  to  have 
8  layers  of  80  turns  each,  giving  a  coil  length  of  8.24  in.,  and  coil 
thickness  of  0.824  in.  The  thickness  will  be  slightly  less,  owing 
to  the  bedding  of  the  layers.  Perfect  bedding  would  give  0.824 
X  0.866  =  0.714  in.  The  value  of  0.75  in.  originally  assumed 
may  therefore  conveniently  be  taken  as  correct. 

The  mean  length  of  the  primary  turn  is  then  LI  =  21.6  in. 
as  previously  calculated. 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  219 


21  6 
Total    length    of    primary  =  -^  X  1280  =  2300   ft.     Total 

primary  resistance  =  2300  X  1.18  =  2.72  ohms. 

This  resistance  deviates  considerably  from  the  value  of  2.4 
ohms  assumed,  but  not  enough  to  warrant  the  choice  of  another 
size  of  wire  for  the  primary. 

Having  determined  the  core  cross-section,  and  the  coils,  an 
assembly  sketch  may  be  made,  as  shown  in  Fig.  163.  Allowing 
0.3  in.  between  the  coils  on  the  two  legs,  the  size  of  the  window  is 


FIG.  163. 

found  to  be  4.24  in.  wide  by  10.75  in.  high.  The  total  core 
height  is  10.75  +  3.05  +  3.05  =  16.85  in.;  total  core  width  is 
4.24  +  3.05  +  3.05  =  10.34  in. 

The  total  volume  of  iron  is  length  X  net  cross-section,  or, 

V  =  (2  X  16.85  -f  2  X  4.24)  X  8.37 

=  42.18  X  8.37  =  354  cu  in., 

which  does  not  compare  very  favorably  with  the  first  assumption 
of  250  cu.  in.,  but  this  is  not  very  important  since  it  should 
be  noted  that  I  calculated  from  core  loss  and  I  calculated  from 


220  ELECTRICAL  ENGINEERING 

the  space  required  for  the  copper  windings  will  generally  not  be  in 
agreement.  We  must  have  sufficient  space  for  the  windings, 
but  I  should  not  be  any  greater  than  necessary.  Therefore, 
unless  we  wish  an  entire  recalculation  of  the  design  based  on 
altered  assumptions  it  is  sufficient  to  accept  the  new  value  of  I 
and  the  attendant  new  value  of  V.  The  mean  length  of  the  flux 
path  is, 

I  =  2  (10.75  +  4.24)  +  27r  X  1.5 
=  29.98  +  9-44  =  39.42  in., 

as  against  29.5  in.  in  the  preliminary  calculations. 

Per  Cent.  Magnetizing  Current  and  Core  Loss.  —  Applying  the 
new  values  of  V  and  I,  we  may  obtain  new  values  for  per  cent. 
magnetizing  current  and  the  core  losses.  Thus, 

amp.  turns  =  MQ  X  I  =  6.5  X  39.42  =  256. 

256 
Max.  exciting  current  =  \/2im  =  TQC     =  0.2. 

' 


Per  cent.  im  =        =  =  0.028. 

/i  5 

Hysteresis  loss,  Wh,  will  be  in  the  ratio  of  the  two  volumes  thus 
far  obtained,  namely;  250  cu.  in.  and  354  cu.  in. 
354 

•  •  Wh  =  250  x  55<2  =  78*2  watts'  and  similarlv  the  eddy 

354 
current  loss  is  We  =         X  24.8  =  35.1  watts,  giving   a  total 


core  loss  of  113.3  watts. 

Efficiency.  —  The  approximate  efficiency  is  then 

input  —  losses 

77    =   --  ;  -  —  —  —  > 

input 
where  the  losses  are: 

Primary  copper  loss  =  52  X  2.72  =68  watts. 

Secondary  copper  loss  =  2  X  502  X  0.012  =    60  watts. 
Core  loss  =  113.3  watts. 

Total  loss  =  241.4  watts. 

*'*  *  =  ~    10  ooo  —  =  °*976  =  97>6  per  cent> 

It  is  seen  that  the  efficiency  is  very  nearly  that  which  was 
assumed  at  the  outset,  so  that  the  variations  in  values,  even  where 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  221 

they  have  been  large,  as  with  volumes  and  length,  have  not  been 
such  as  to  produce  any  considerable  effect. 

TT  »  copper  loss  . 

However,  the  ratio  of    ^ron  |oss     nas  decreased,  the  former 

now  constituting  only  53  per  cent,  of  the  total  loss,  instead  of  60 
per  cent,  as  at  first  assumed. 

There  is,  of  course,  nothing  hard  and  fast  about  these  relations 
as  it  is  impossible  to  say  just  how  the  transformer  will  be  oper- 
ated from  day  to  day.  If  it  is  desired  to  approximate  more 
closely  to  the  original  assumptions  of  losses  and  efficiency,  it  will 
be  necessary  to  go  back  to  the  beginning  and  choose  from  among 
the  numerous  variables,  let  us  say,  another  value  of  turns  and 
another  value  of  the  flux  density. 

However,  in  the  present  instance,  the  values  so  far  obtained 
will  be  regarded  as  satisfactory,  and  then  there  remains  only 
to  determine  the  regulation,  heating  and  cost  of  material,  to  see 
if  these  also  will  be  satisfactory. 

Regulation. — In  Chap.  XXVI,  p.  182,  the  regulation  of  a  trans- 
former was  found  to  be 

Reg.  =  ^  "  1> 
where    EI  =  V  E22  +  2Ez(i2rQ  +  0.54n>  -  i'&o  +  0.5imz0) 


approximately. 

Of  these  quantities,  EI  =  2000  volts,  assumed  impressed  on 
the  primary,  iz  =  72  =  5  amp.  =  the  energy  component  of  the 
load  at  unity  power  factor  (assumed)  and  referred  to  the  primary 
basis.  On  this  assumption,  the  wattless  component  of  the  load 
current,  i'z  =  0. 

4  =  energy  component  of  the  exciting  current. 

To  obtain  4,  we  have:  core  loss  =  Eih  =  113.3  watts. 

1100 

•"• ih  =  ^  =  °-0566  amp*' 

and  

too  =  Vi^+42  =  V(H412  +  0.05662  =  0.152  amp. 
r0  =  fli  +  #2  =  2.72  +  0.009  X  400  =  2.72  +  3.6  =  6.32. 
To  determine  XQ,  the  combined  leakage  reactance  of  primary 

and  secondary,  we  have 

XQ    =    27T/  (Li   +  1/2)    =    27T/I/0. 

LI  and  L2  may  now  be  calculated  by  the  help  of  equations, 
p.  204,  Chap.  XXX,  but  each  must  be  done  separately  since  the 


222  ELECTRICAL  ENGINEERING 

two  primary  coils  are  in  series  while  the  secondary  coils  are  in 
parallel.     We  have 


Referring  to  Fig.  155,  the  constants  in  these  equations  are 
readily  evaluated.     Thus,  we  have, 
Ni  =  640;  N*  =  64;  I  =  10.75, 
mi  =  mean  length  of  primary  turn  =  21.6  in., 
mz  =  mean  length  of  secondary  turn  =  15.4  in., 
mz  =  mean  length  of  gap  between  coils  =  18  in., 

a  =  secondary  coil  thickness  =  0.44  in., 

b  =  distance  between  coils     =  0.39  in., 

c  =  primary  coil  thickness     =  0.75  in. 
Supplying  these  values, 

fc  -  64  X  10-9     41  X  1Q4   2L6X°'75       18X°'39 
Ll  ~ 


10.75  3 

=  0.00244  [5.4  +  3.51]  =  0.02175  henry 


=  6.1  X  10-6  [2.26  +  3.51]  =  0.0000352  henry, 
where  L2  is  the  actual  secondary  inductance. 

Referred    to    the    primary,    L2  =  0.0000352  X  400  =  0.0141 
henry. 

Lo  =  Li  +  L2  =  0.02175  +  0.0141  =  0.03588  henry 
and 

Zo  =27r/Lo  =  377  X  0.03588  =  13.53  ohms. 
Supplying  all  the  values  into  the  formula  for  Ei,  we  have, 
Ei  =  2000  = 


X  6.32+0.5  X  0.0566  X  6.32  +  0.5  X  0.141  X  13.53) 
4  X  106  =  #22  +  2#2(31.6  +  0.179  +  0.95)  = 


V  2000 

.'.    E2  =  1968  volts,  and  regulation  —  —  —  1  «•  —  —  —  1  = 

1.016  —  1  =  0.016  =  1.6  per  cent,  for  full  non-inductive  load. 

Heating.  —  The  total  radiating  surface  of  each  primary  coil  is 
found  by  calculation  to  be  388  sq.  in.  Therefore,  the  watts  per 
square  inch  which  must  be  radiated  from  the  primary  coil  are 


.  0.0875. 

388  sq.  in. 


GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  223 

Similarly,  the  area  of  the  secondary  coil  radiating  surface  is  340 
sq.  in.     The  watts  per  square  inch  that  must  be  radiated  are 

-«•««• 

The  radiating  surface  of  the  core  is  about  400  sq.  in.     Therefore 
watts  per  square  inch  that  must  be  radiated  are 

__  113.3  watts 
'   400  sq.  in. 

Watts  per  square  inch  serve  as  an  empirical  guide  by  which  it 
may  be  determined  satisfactorily  whether  the  design  is  sufficiently 
liberal  to  permit  of  dissipation  of  the  heat  without  undue  rise  of 
temperature  of  any  part.  In  general  a  loss  of  0.4  watt  per  sq.  in. 
of  surface  of  the  coils  and  core  is  quite  satisfactory. 

In  designing  the  case,  however,  about  0.15  watts  per  sq.  in. 
only  should  be  allowed.  In  the  transformer,  then,  since  the 
entire  loss  in  watts  must  be  radiated  from  the  case  we  should 
need  an  area  of 

241.3  watts  .       .  .,,   ,,       ., 

fT-^= =  1610  sq.  m.,  in  contact  with  the  oil. 

U.lo 

Weight  and  Cost  of  Material. — The  core  volume  has  been 
found  to  be  354  cu.  in.  At  0.28  Ib.  per  cu.  in.  the  core  weight  is 

354  X  0.28  =  99  Ib. 
Cost  of  core  at  3.5  c.  per  Ib.  is 

99  X  0.035  =  $3.46. 

The  primary  copper  volume  is  length  X  section, 
=  2300  X  12  X  0.00815  =  225  cu.  in. 
Secondary  copper  volume  is 

82  X  12  X  0.0655  X  2  =  129  cu.  in. 
Total  volume  of  copper  is  225  +  129  =  354  cu.  in. 
Weight  of  copper  at  0.32  Ib.  per  cu.  in.  is 

354  X  0.32  =  113.4  Ib. 
Cost  of  copper  at  16c.  per  Ib.  is 

113.4  X  0.16  =  $18.15. 
Total  cost  of  iron  and  copper  is  $3.46  +  $18.15  =  $21.61 

Of  course,  such  a  calculation  of  cost  has  comparative  merit 
only,  as  it  does  not  include  labor  or  such  materials  as  insulation, 
oil  and  case. 


224 


ELECTRICAL  ENGINEERING 


Summary  of  data  of  10-kw.,  60-cycle  2000-100-volt  core-type 
distributing  transformer. 


- 

High  side 

Low  side 

Kilowatts                                   

10 

Frequency 

60 

Ratio  of  transformation  

20:1 

Volts                                           

2  000 

100 

Arnp6r6s 

5 

100 

Window  dimensions,  in  

10%  by  424 

Total  width  of  iron!!  in  
Total  height  of  iron,  in  

10.34 
16.85 

Depth  of  lamination,  in  
Electrical 
Number  turns  in  series 

3.05 

1  280 

64 

Section  of  conductor  
Amperes  per  square  inch 

0.00815 
614 

0  .  0655 
763 

Number  of  coils 

2 

2 

Connection  of  coils.  ... 

Series 

Parallel 

Width  of  coil 

0  75 

0  44 

Height  of  coil  

8  24 

10.6 

Number  turns  per  coil 

640 

64 

Mean  length  of  turn  
Resistance  of  circuit  at  65°C 

21.6 
2  72 

15.4 
0.009 

Magnetic 
Total  maximum  flux  
Effective  core  section,  sq.  in. 

586,000 

8  37 

Effective  core  length,  in  
Core  density  
Effective  core  ampere-turns  

39.42 
70,000 
180.5 

Magnetizing  current  
Thermal 
/'flloss  

0.141 

68 

60 

Radiating  surface  of  coil  

388 

340 

Watts  per  square  inch 

0  0875 

0  0885 

Core  loss  

113  3 

Radiating  surface  of  core  
Watts  per  square  inch 

400 
0  28 

Total  loss,  full-load  

241  3 

External  radiating  surface  
Watts  per  square  inch 

1,610 
0  15 

Efficiency  and  regulation 
Per  cent,  core  loss,  full-load  
Per  cent,  copper  loss,  full-load  . 

1.15 
1  3 

Per  cent,  efficiency,  full  load 

97  6 

GENERAL  PRINCIPLES  OF  TRANSFORMER  DESIGN  225 


High  side 

Low  side 

Per  cent,  magnetizing  current  
Per  cent,  resistance  

2.8 
1  58 

Per  cent,  reactance.  .  . 

3  38 

Per  cent,  regulation 

1  6 

Weight  and  cost 
Copper,  pounds  

113  4 

Iron,  pounds 

99  0 

Pounds  copper  per  kilowatt  

11  34 

Pounds  iron  per  kilowatt  . 

9  9 

Cost  of  copper  at  16c  
Cost  of  iron  at  3.5c  
Total  cost. 

$18.15 
$3.46     . 

$21  61 

Cost  per  kilowatt 

$2  16 

Having  now  developed  the  general  principles  and  procedure  in 
transformer  design,  it  is  desirable  that  the  student  should  carry 
through  the  calculations  for  some  assigned  machines.  Trans- 
formers of  different  capacity  may  be  assigned  to  the  students  of  a 
section,  each  student  being  required  to  complete  the  calculations 
for  both  60  cycles  and  25  cycles,  tabulating  the  specifications  and 
making  sketches  to  scale  of  the  core  and  windings. 

Many  of  the  finer  points  in  design  are  omitted  here,  since  the 
principles  are  the  primary  interest.  For  practical  designing  the 
fact  that  experience  is  a  factor  of  the  greatest  importance  should 
always  be  remembered  by  the  student  who  is  attempting  to 
master  the  practical  aspects  of  the  subject. 

To  assist  in  the  further  study  of  the  principles  of  transformer 
design,  the  following  suggestive  questions  are  added. 

1.  Find  regulation  at  80  per  cent,  power  factor. 

2.  Why  are  less  volts  per  turn  used  with  lighting  than  with 
power  transformers? 

3.  If  the  core  loss  is  too  great,  how  may  it  be  reduced? 

4.  What  relation  does  per  cent,  exciting  current  have  to  core 
loss,  copper  loss,  efficiency  and  regulation? 

5.  How  may  per  cent,  exciting  current  be  reduced? 


15 


CHAPTER  XXXII 

COMBINATIONS  IN  MULTIPHASE  TRANSFORMER 
SYSTEMS 

When  the  primary  of  a  transformer  is  connected  to  a  source 
of  e.m.f  .  the  following  equation  relating  the  impressed  e.m.f  .  cur- 
rent, resistance  and  inductance  obviously  obtains 

e  =  ri  +  ^  (Li). 

The  drop,  ri,  is  small,  being  perhaps  5  per  cent,  of  1  per  cent. 
of  the  normal  voltage,  if  the  exciting  current  is  5  per  cent,  of 
normal  current  and  the  resistance  drop  at  full-load  is  1  per  cent. 
Thus  the  e.m.f.  consumed  by  the  transformer  counter  e.m.f.  is 
approximately 


If  the  transformer  were  merely  a  coil  having  an  air  core,  the 
inductance  would  be  constant,  and  the  induced  voltage  would  be 

.  di 


If  i  were  a  sine  wave  of  current,  e  would  also  be  a  sine  wave 
displaced  90°  behind  i.  However,  with  iron  cores,  as  with  trans- 
formers, the  inductance  is  not  constant,  but  is  a  function  of  the 
current  i.  Hence 

di   .    . 


and  if  i  is  a  sine  wave  of  current,  the  counter  e.m.f.  of  self-induc- 
tion is  no  longer  a  sine  wave.  Similarly,  if  a  sine  wave  e.m.f. 
is  impressed  on  the  transformer,  the  current  will  not  have  the 
sine  shape,  but  will  be  made  up  of  fundamental,  3d,  5th,  etc., 
harmonics. 

In  Chap.  XXVIII,  the  characteristic  wave  of  exciting  current 
with  a  sine  wave  e.m.f.  impressed  was  determined.  On  analyzing 
this  wave  by  FOURIER'S  series  as  indicated  in  problem  78,  it  is 
found  to  consist  of  a  fundamental  and  triple  with  higher  har- 

226 


MULTIPHASE  TRANSFORMER  SYSTEMS       227 


monies  of  lesser  amplitudes.  The  presence  of  the  triple-fre- 
quency wave  is  an  important  feature  in  the  exciting  current  of 
every  iron-cored  transformer  operating  with  impressed  sine  wave 
e.m.f. 

If,  however,  the  triple  frequency  wave  is  suppressed  in  some 
way,  as  is  often  the  case  in  three-phase  systems,  so  that  the  ex- 
citing current  is  of  sine  shape  (neglecting  small  higher  harmonics), 
the  induced  voltage,  and  consequently  the  terminal  voltage,  will 
not  be  of  sine  shape,  but  will  have  the  characteristic  form  shown 
on  p.  194. 

The  transformer,  of  course,  must  generate  its'  own  counter 
e.m.f.  or  the  induced  voltage,  and  hence  may  be  regarded  as  a 
generator,  electrical  energy  being  supplied  to  it  instead  of  me- 
chanical energy. 

So  far  the  transformer  has  been  dealt  with  as  a  single  unit. 
It  is  common  practice,  however,  to  group  transformer  units  in 
various  ways  so  that  they  shall  serve  as  group  units  in  the  trans- 
mission and  distribution  of  energy  in  systems  other  than  the 
single-phase  system. 

The  Three-phase  System. — While  two-phase,  four-phase  and 
six-phase  systems  are  used  to  some  extent  and  under  certain  con- 
ditions, yet  the  three-phase  ^ 

system  in  its  various  forms 
is  far  more  important  than  all 
of  these.  Its  study  forms  a 
basis  for  the  development  of 
any  multiphase  theory.  The 
principles  of  two-phase  and 
three-phase  working  from  the 
standpoint  of  the  alternator 
are  explained  in  Chap. 
XXXV.  In  the  present  in- 
stance, three-phase  will  be 
dealt  with  in  reference  to  the 
transformer  alone. 

Consider  three  similar  transformers,  A,  B  and  C,  receiving 
current  from  three  sources  of  simple  sine  waves  of  e.m.f.  (Fig.  164). 
Let  the  voltage  impressed  on  A  be  et  =  EI  sin  6,  that  on  B, 
e*  =  El  sin  (0  +  120°),  that  on  C,  e3  =  EI  sin  (6  +  240°).  In 
each  case  the  arrows  in  the  figure  indicate  outgoing  and  return 
wires. 


FIG.  164. 


228  ELECTRICAL  ENGINEERING 

Neglecting  higher  harmonics  the  currents  flowing  will  be,  re- 
spectively 

ii  =  /i  sin  (0  —  0) 

iz  =  I,  gin  (6  +  120°  -  0) 

iz  =  /!  sin  (0  +  240°  -  0). 


If  the  transformers  are  so  arranged 
that  the  return  currents  shall  flow 
through  the  same  wire,  as  in  Fig.  165, 

the  value  of  the  current  in  this  return 

FIG.  165.  wire  will  be 

in  =  ii  +  iz  +  iz  =  Ii  (sin  (0  -  0)  -f  sin  (0  +  120°  -  0) 

+  sin  (6  +  240°  -  0). 

The  student  should  prove  that  this  current  is  zero. 

He  should  prove,  also,  that  if  there  is  current  of  triple  frequency 
flowing  in  the  lines,  the  triple  frequency  current  in  the  fourth  or 
neutral  wire  will  be  three  times  that  in  any  line. 

Problem  79. — Let  the  three  line  currents  be  given  by  the  equations: 
ii  —  Ii  sin  (0  —  $0  +  /a  sin  (30  —  <£s)  +  It  sin  (50  —  $5)  +   .    .    .    . 
it  =  /i  sin  (0  +  120°  -  00  +  73  sin  [3(0  +  120°)  -  03j 

+  h  sin  [5  (0  +  120°)  -  <fa]  +  .    .    . 
iz  =  /i  sin  (0  +  240°  -  00  +  h  sin  [5  (0  +  240°)  -  0(]  + 

/ssin  [5(0  +240°)  -««]  +  .    .    . 
Prove  that  the  current  in  the  neutral  wire  will  be 
in  =  3[/3  sin  (30  —  <£3)  +  1 9  sin  (90  —  <£9)  +  /i6  sin  (150  —  #12)  +  J.    •    •> 

that  is,  the  current  in  the  neutral  wire  is  three  times  the  sum  of  all 
odd  harmonics  which  are  multiples  of  three  which  are  present  in 
any  line  wire,  all  other  harmonics  becoming  zero  in  the  neutral. 

Voltage  Waves  in  Three-phase,  Four-wire  System. — The  neu- 
tral wire  serves  to  make  the  system  virtually  three  single-phase 
systems  instead  of  a  three-phase  system  having  peculiarities  of 
its  own.  Thus  if  a  sine  wave  e.m.f.  is  impressed  on  each  trans- 
former, the  e.m.f.  between  lines  is  the  vector  sum  of  any  two 
of  these  and  is  also  a  sine  wave. 

Since  there  is  no  current  of  fundamental  frequency  in  the 
neutral  wire,  there  is  no  necessity  of  having  the  wire  there.  Its 
absence  will  not  be  the  occasion  for  any  interruption  in  the  circuit. 
However,  since  the  circuit  of  the  higher  harmonics  has  been  inter- 
rupted, it  becomes  of  importance  to  study  higher  harmonic  effects 
in  connection  with  three-phase  and  with  any  other  systems. 
Transformers  so  arranged  with  or  without  the  neutral  wire  are 


MULTIPHASE  TRANSFORMER  SYSTEMS        229 

said  to  be  Y-connected.  If  the  circuits  were  unbalanced  the 
situation  would  be  somewhat  different,  as  will  be  discussed  later. 
For  the  present,  however,  it  is  sufficient  to  see  that  a  three-phase 
circuit  may  be  composed  of  only  three  wires,  each  representing 
the  outgoing  wire  of  one  of  the  phases. 

Three-phase,  Y-connected  Transformers.— Let  it  be  assumed 
that  the  three,  so-called,  phase  voltages  are 

OA  =  BI  =  El  sin  B  +  Es  sin  (30  +  a), 

OB  =  62  =  El  sin  (0  +  120°)  +  Ez  sin  (3[0  +  120°]  +  a), 

OC  =  e3  =  #1  sin  (0  +  240°)  +  E3  sin  (3[0  +  240°]  +  a), 

these  voltages  being  represented  vectorially  in 
Fig.  166.  To  find  the  line  voltage  AB.  Evi- 
dently this  is 


eAB  =  AO  +  OB  =  - 


since  it  is  taken  in  direction  from  A  to  B. 
Directions  from  0  outward  are  taken  as  posi- 
tive. Therefore 

CAB  =  -  EI  sin  0  -  Ez  sin  (30  +  «)  +  El  sin  (0  +  120°)  -f  E3 

sin  (3[0  +  120°]  +  a) 

=  #i[sin  0  cos  120°  +  cos  0  sin  120°  -  sin  0] 
+  Et[sm  (30  +  a)  -  sin  (30  +  a)]. 

The  last  term  vanishes  since  sin  (3[0  +  120°]  +  a)  =  sin  (30  -f 
360°  +  a)  =  sin  (30  +  a)  and  eAB  =  1.73  EI  sin  (0  +  150°). 

The  student  should  prove  this  by  performing  the  intermediate 
operations. 

Thus,  it  is  seen  that  in  a  balanced  three-phase  Y-connected  sys- 
tem, a  triple  frequency  e.m.f.  cannot  exist  in  the  voltage  between 
the  lines.  The  same  will  be  shown  to  be  true  also  for  what  is 
called  the  A-connection.  This  does  not  mean,  as  stated,  that 
there  can  be  no  triple  frequency  e.m.fs.  in  the  phase  windings, 
but  simply  that  they  cannot  be  between  the  lines. 

Likewise,  the  other  line  voltages  are: 

eBC  =  -  1.73  EI  sin  (0  +  90°) 
CCA  =  1-73  E!  sin  (0  +  30°). 

These  voltages  are  represented  in  Fig.  167  (a)  and  (6),  which 
give  two  ways  of  representing  the  same  thing.  The  line  and 
phase  voltage  relations  may  also  be  shown  graphically,  as  in 


230 


ELECTRICAL  ENGINEERING 


167. 


— .e 


Fig.  168.     Here,  as  in  the  equation  for  e^B)  OB  is  combined  with 
OA  reversed. 

Carrying  out  the  same  process  by  which  the  assumed  triple- 
frequency  voltages  in  the  line  were  eliminated,  it  could  also  be 
found  that  any  higher  harmonics  which 
were  multiples  of  3,  as  9th,  15th,  21st, 
etc.,  would  vanish.  Even  harmonics  are 
of  necessity  absent  if  the  waves  are  sym- 
metrical. Thus  there  remain  only  the 
5th,  7th,  llth,  13th,  17th,  19th,  etc., 
which  could  exist  in  the  lines. 
In  the  three-phase  Y-connection,  the  triple  frequency  voltages 
in  the  phases  are  all  in  time-phase  with  each  other,  and  the  phase 
therefore  acts  like  three  circuits  in  parallel.  Their  extremities 
could  be  joined  without  causing  any  triple  frequency  current  to 
flow.  If  the  neutral  point,  0,  be  connected  to  ground,  the  triple 
frequency  in  the  phases  would  cause  all  three  transmission  lines 
to  oscillate  just  as  would  be  the  case  with  a  single-phase  line  one 
side  of  which  was  grounded. 
In  this  case  the  three  lines  cor- 
respond to  the  ungrounded  side 
of  the  single-phase  line. 

Three-phase  A-connected 
Transformers. — W  hen  three 
transformers  are  so  connected  as 
to  form  a  closed  circuit,  there 
are  two  facts  in  connection  with 
their  operation  which  are  of 
great  interest,  namely:  (1) 
there  can  be  no  circulating  cur- 
rent of  fundamental  frequency 

in  the  windings;  and  (2)  there  always  flows  in  the  windings  a 
current  of  triple  frequency  or  an  odd  multiple  of  triple  frequency. 
In  proof  of  the  first  fact  let  the  phase  voltages  be  assumed,  as 
before, 

Ci  =  EI  sin  0  +  Ez  sin  (3  0  +  a) 

e2  =  El  sin  (0  +  120°)  +  #3  sin  [3(0  +  120°)  +  a] 

e3  =  Ei  sin  (0  +  240°)  +  E*  sin  [3(0  +  240°)  +  a)] 

Adding  the  fundamental  components, 

El  [sin  6  +  sin  (  0+  120°)  +  sin  (0  +  240°)] 


FIG.  168. 


MULTIPHASE  TRANSFORMER  SYSTEMS        231 


FIG.  169. 


=  Ei  [sin  0  +  sin  0  cos  120°  +  cos  0  sin  120°  +  sin  0  cos  240°  + 

cos  0  sin  240°] 
=  Ei  [sin  0  +  sin  0  X     (-  0.5)  +  cos  0  X  (0.866)  +  sin  0  X 

(-  0.5)  +  cos  0  X  (-  0.866)] 
=  EI  [sin  0  —  sin  0)  =  0. 

Thus,  if  there  is  no  e.m.f.  of  fundamental  frequency  acting  in 
the  closed  winding,  there  can  be  no  current  of  fundamental 
frequency  circulating  in  it. 

In  proof  of  the  second  fact,  adding  the  triple-frequency  com- 
ponents gives 

#3  sin  (30  +  a)  -f  Ez  sin  [3(0  +  120°)  +  a]  +  #3  sin  [3(0  + 
240°)  +  a)  =  SE's  sin  (30  +  a). 

Thus  if  the  delta  is  open  at  one  point  (Fig.  169)  the  triple  voltage 
across  the  opening  is  three  times  the  triple-frequency 
voltage  of  one  phase.  Similarly,  with  a  Y-connec- 
tion  with  neutral  point  grounded,  the  triple-fre- 
quency current  flowing  into  the  ground  is  three  times 
the  triple-frequency  current  of  one  phase.  When 
the  neutral  is  grounded,  it  is  no  longer  necessary  to  regard  the 
system  as  three-phase,  but  it  may  be  considered  as  three  single 
phases  having  a  common  return,  just  as  with  the  three-phase  four- 
wire  system  already  discussed. 

The  triple-frequency  current  is 
then  perfectly  free  to  flow  in  each 
line  wire,  returning  by  way  of  the 
neutral,  whereas  without  the  neutral, 
the  triple-frequency  current  cannot 
exist. 

To  find  the  relation  between  line  current  and  phase  current 
in  a  A-connected  system. 

Let  the  direction  of  the  phase  currents  be  assumed  as  indicated 
in  Fig.  170  where 

ii  =  7i   sin   0  +  73   sin    (30  +  a) 

iz  =  h   sin    (0  +  120°)  +  73  sin  [3(0  +  120°)  +  a] 

i3  =  1 1  sin  (0  +  240°)  +  73  sin  [3(0  +  240°)  -f  a] 

Taking  the  direction  of  the  arrows  as  positive,  the  funda- 
mental line  current, 

*A  =  ii  ~  i*  =  1 1  (sin  0  -  sin  (0  +  120°))  =  1.73/isin  (0  -  30°) 


FIG.  170. 


232  ELECTRICAL  ENGINEERING 

This  relationship  is  similar  to  that  of  the  voltages  for  Y-connec- 
tion.  Similarly,  also,  there  can  be  no  triple-frequency  current  in 
the  line. 

Voltage  Waves  with  Y-connected  Transformers. — In  problem 
76,  was  assumed  a  sine  wave  of  exciting  current.  This  is  approx- 
imately the  case  with  the  three-phase  Y-connection  since  there 
can  be  no  triple-frequency  current  in  the  line  or  phase. 

The  phase  voltage  must  then  look  like  that  of  Fig.  151.     The 

line  voltage  as  previously  seen  will 
be  a  combination  of  two  phase 
voltages,  one  of  which  is  reversed, 
as  in  Fig.  171. 

(The  depression  in  the  line  volt- 
age wave  is  not  actually  as  deep  as 
would  appear  from  using  sine 
waves  of  magnetizing  current.) 

If  the  generator  develops  a  sine  wave  e.m.f .  and  the  transformer 
counter  e.m.f.  is  much  distorted,  due  to  the  hysteresis  loop  effect, 
then  the  difference  between  these  two  waves  must  be  taken  up  by 
drops  along  the  lines  and  in  the  apparatus. 

Problem  80. — Given  three  A-connected  transformers.  Make  a  picture 
of  the  e.m.f.  and  compare  it  with  that  of  a  single-phase  circuit.  What  is 
the  shape  of  the  wave  of  phase  current?  What  is  the  shape  of  the  wave  of 
line  current?  Show  also,  by  a  sketch  that  the  sum  of  two  exciting  current 
waves  of  a  three-phase  A-connected  system  makes  nearly  a  sine  wave,  the 
triple  frequency  vanishing.  How  does  the  core  loss  in  this  system  compare 
with  that  of  three  single-phase  circuits? 

Problem  81. — Given  three  Y-connected  transformers.  The  line  current 
can  contain  no  triple  harmonics  but  only  1st,  5th,  7th,  llth,  etc.,  harmonics. 
The  phase  voltage,  however,  has  a  large  triple  harmonic.  The  line  voltage 
is  a  combination  of  two-phase  voltages.  What  is  the  ratio  of  the  phase 
voltage  to  the  line  voltage?  Is  it  58  per  cent.?  Evidently  it  is  higher,  as 
the  phase  voltage  contains  also  the  triple- 
frequency  voltage.  How  does  the  core  loss 
of  this  system  compare  with  that  of  the  A- 
connected  system  and  with  the  single-phase? 

The  flux  wave  is  flat,  since  there  is  no 

triple-frequency    current.      Therefore     the  pIGi  172. 

maximum  value  of  flux  is  less,  and  the 
core  loss  is  less  (by  about  30  per  cent.),  than  with  sine  waves  of  flux. 
Moreover,  the  exciting  current  is  less  because  the  max.  value  of  the  flux 
density  is  less. 

With  open  delta  connection  what  will  the  voltmeter  read?  Evidently 
three  times  the  triple-frequency  voltage  per  phase.  Why?  With  the 


MULTIPHASE  TRANSFORMER  SYSTEMS        233 

neutral  wire  connected  in  a  Y-system  as  in  Fig.  172,  what  would  be  the 
current  in  the  neutral? 

As  has  been  pointed  out,  this  is  no  longer  a  real  three-phase  system,  but 
three  single  phases  in  which  the  neutral  wire  is  common  to  all  the  phases. 

Evidently  the  triple-frequency  currents  can  flow  in  each  phase,  and  since 
they  are  all  in  time-phase  with  each  other  the  neutral  will  carry  three  times 
the  triple  harmonic  current  of  each  phase. 

What  then,  will  be  the  effect  on  the  core  loss  of  connecting  in  the  neutral, 
as  compared  with  leaving  it  out? 

These  problems  are  stated  in  such  a  way  as  to  form  the  basis  for  a  fairly 
complete  discussion,  on  the  part  of  the  student,  of  the  effects  which  would 
be  produced  by  the  different  ways  of  connecting  the  transformer. 

Three-phase  Transformers. — A  natural  development  in  the 
use  of  three  single-phase  transformers  for  three-phase  work  is  the 
substitution  therefor  of  a  single  three-phase  transformer. 


(d) 


Po  Po 

' 

po         (e)       Po 
FIG.  173. 


Let  there  be  three  cores  of  laminated  iron,  symmetrically  placed, 
connected  by  legs,  each  core  having  on  it  the  windings  of  one 
phase.  This  may  be  done  as  in  Fig.  173,  a,  b.  How,  then,  should 
the  sectional  area  of  the  core  be  calculated?  This  should  evi- 
dently be  done  in  the  regular  way  since  each  leg  has  its  coil,  and 
must  have  its  flux  set  up  by  the  coil.  The  yoke,  however,  cor- 
responds to  a  A-connection,  and  the  flux  in  any  leg  of  the  yoke  is 

—=  =  58  per  cent,  of  the  flux  in  any  core. 


The  yoke  may  also  be  formed  as  a  Y-connection  (Fig.  173,  c,  6), 
in  which  the  flux  in  any  branch  of  the  Y  is  the  same  as  that  in 
any  leg. 


234 


ELECTRICAL  ENGINEERING 


In  practice,  however,  it  is  common  to  employ  a  form  such  as 
Fig.  173,  dy  in  which  there  are  three  equal  legs  carrying  the  coils 
and  the  yoke  is  straight  across  the  top  and  bottom.  The  whole 
core  is  built  up  of  laminations,  which,  except  for  the  diffi- 
culty of  placing  the  coils,  could  be  of  one  piece.  In  Fig.  173,  d, 
the  flux  paths  are  outlined  by  dotted  lines,  and  it  is  evident  that 
so  far  as  the  magnetic  core  is  concerned,  coils  1  and  3  are  sym- 
metrical with  respect  to  each  other,  while  coil  2  is  unsymmetrical 
with  respect  to  1  and  3. 

Assume  the  reluctance  of  one  leg  to  be  p,  and  the  reluctance 
of  one  section  of  the  yoke  to  be  po. 

Then  there  may  be  constructed  an  analo- 
gous electric  circuit,  as  in  Fig.  173,  e.  This 
gives  the  magnetic  circuit  which  is  supplied 
with  flux  by  the  m.m.f.  of  coil  1,  on  leg  1. 
It  may  be  simplified  to  Fig.  174,  in  which 


. 

p'       p      p  +  2p0 


FIG.  174. 


p'  = p(p 


2p0) 


2(p  +  po) 
The  total  reluctance  of  the  magnetic  circuit  of  coil  1,  is  thus 


2p0 


(3p  -f  2Po) 


^- 


2P: 


"2(p  + 

It  is  also  evident  that  pi  =  p3. 
For  p2,  the  circuit  may  be  considered  as 
made  of  two  parallel  paths,  as  in  Fig.  175. 
Here, 

1  p«  ; 

P2  =  2  (^P  +  2p0).  FIG.  175. 

Thus  the  relative  reluctances  of  the  two  circuits  are 


P2 
Pi 


PO 


2p 


In  a  good  transformer,  the  ratio  of  height  to  width  of  the 
window  is  from  4  to  8:1;  the  average  is  about  6:1. 
.'.  P  is  from  4  to  8  times  as  large  as  p0. 


MULTIPHASE  TRANSFORMER  SYSTEMS        235 


Assuming  p  =  6p0, 

P2  _  Tpo  _  7 
Pi       8po       8 

.'.  p2  has  87K  per  cent,  as  great  a  value  as  Pl,  and  the  exciting 
current  in  the  middle  coil  is  from  80  per  cent,  to  90  per  cent,  of 
that  in  the  outside  coils. 

Suppose  it  were  necessary  to  have  equal  exciting  current  in  all 
the  coils.  This  could  be  accomplished  by  reducing  the  section 
of  the  middle  leg. 

But,  in  this  case,  the  core  loss  would  be  unbalanced,  for  it  is 
approximately  proportional  to  the  square  of  the  flux  density 
which  would  be  increased  in  the  middle  leg. 

Therefore,  there  must  be  some  unbalancing.  In  practice  all 
parts  are  made  of  equal  section,  including  the  yoke.  It  is  there- 
fore easy  to  calculate  the  saving  in  material  over  three  single- 
phase  transformers. 

Question. — Considering  wave  shapes  as  discussed  above  if  the 
coils  are  connected  Y,  will  there  be  a  triple-frequency  voltage? 

It  has  been  shown  that  the  triple-fre- 
quency currents  are  in  time-phase  in  the 
different  phases.  Hence  they  produce 
fluxes  in  time-phase  with  each  other.  These 
then  neutralize  each  other  or  pass  around 
through  the  air  which  makes  them  very 
weak.  The  induced  triple-frequency  volt- 
ages are  therefore  very  small.  Thus  a  three-phase  Y-connected 
transformer  acts  like  three  choking  coils  as  far  as  the  triple-fre- 
quency current  is  concerned.  Their 
flux  paths  being  largely  in  air,  the 
hysteresis  loops  are  very  thin,  causing 
small  distortion.  Therefore,  core- 
type  three-phase  transformers  behave 
much  like  three  A-connected  single- 
phase  transformers  as  regards  triple- 
frequency  harmonics. 

Shell-type     Three-phase     Trans- 
formers.— These  could  be  made   by 
FIG.  177.  placing  three  single-phase  shell- type 

transformers  one  on  the  other.  In  such  a  case,  the  leg  with  the 
coil  has  a  width,  a,  while  the  other  legs  have  widths  a/2  (Fig. 
177). 


FIG.  176. 




!     ! 

236 


ELECTRICAL  ENGINEERING 


The  combined  intermediate  sections  or  widths,  6,  could  be  re- 
duced to  —n~  a,  since  the  fluxes  differ  by  30°  in  time-phase,  in 

adjacent  intermediate  sections.  This  is  seen  to  be  the  case  by 
noting  the  dotted  lines  in  the  figure.  Arrows  indicate  what  may 
be  called  the  positive  direction  of  the  flux  and  these  directions 
are  opposite  in  the  adjacent  intermediate  sections. 

If  now  the  middle  coil  is  reversed,  the  positive  direction  of  the 
flux  in  the  middle  transformer  is  reversed,  and  the  flux  phases  in 
intermediate  adjacent  sections  have  a  60°  relation.  The  total 

flux  in  these  sections  is  therefore 
exactly  the  same  as  that  in  the  out- 
side section,  and  required  width  of 
section  b  is  also  that  of  the  outside 
section  namely,  a/2.  Transformers 
are  therefore  designed  as  in  Fig.  178, 
with  all  width  dimensions  a/2,  except 
the  middle  legs  which  have  the 
width,  a. 

To    prove    that   in   reversing   the 
middle  coil  the  flux  produced  is   in 
amount  the  same  flux  as  when  the 
coil  is  not  reversed. 
The  flux  produced  by  coil  1  is  $  sin  0.     Flux  produced  by  coil 
2,  reversed,  is  -  $  sin  (0  +  120°).     The  flux  due  to  the  two  coils 
is  then 

$  [-  sin  0  -  sin  (B  +  120°)] 
=  3>  [-  sin  0  -  sin  6  cos  120°  -  cos  6  sin  120°] 
-  $  [0.5  sin  0  +  0.866  cos  0]  =  -  &  sin  (0  +  a) 

where  a.  =  30°. 

Problem  82. — Discuss  the  wave  shapes  of  three-phase  transformers. 
Show  that  in  the  core  type,  it  makes  very  little  difference  whether  the  neutral 
is  connected  or  not. 

Show  that,  in  the  shell  type,  the  waves  are  essentially  the  same  as  those 
of  three  single-phase  transformers. 

Open  Delta  Transformer  Connection. — If  one  of  three  delta- 
connected  transformers  is  disabled  it  is  possible  to  operate  at 
reduced  output  with  the  remaining  two,  connected  as  shown 
in  Fig.  179. 

The  following  is  a  comparison  of  the  use  of  two  transformers 


FIG.  178. 


MULTIPHASE  TRANSFORMER  SYSTEMS        237 

and  three  transformers  when  the  power  delivered  is  assumed 
equal  in  the  two  cases. 

A  < 

Current  in  transformer /„,  — 7=  / 

V§ 

Voltage  across  transformer. . .        Ep,  E  E 

TJIT 

Rating  of  each  transformer. . .  El 

v  3 

ETT 

Rating  of  installation 3  —/=  =  \/3EI  2EI 

Ratio   of   transformer   capacity  =  -~—   in   favor   of   the  three 

« 

transformers.     However,  it  may  be  cheaper,  in  a  given  initial 

installation,  to  buy  two  large  transformers     r         

than  three  small  ones.  i 

Now,    assume    a    three-transformer  in-     / 


stallation  in  which  one  transformer    has  \ 

been    disabled.     How    much    should    the 
load  be  reduced  to  give  normal  operation   -  -- 
of  the  remaining  two  on  open  delta? 

The  line  current  must  evidently  be  reduced  in  the  ratio  —  -p 

v  3 

since  the  line  current  and  the  phase  current  are  now  the  same. 

The  output,  which  was  -\/3EI,  therefore  becomes  -\/3E  —7=  =  El. 

V  3 

ETT 

Therefore  the  ratio  of  outputs  is  —  x-       =  0.58  or  less  than  the 

V3-M       ___ 


ratio  of  transformer  capacity  which  is  2/3  or  0.0SB. 

b  Two  transformers  are  frequently 

used  both  for  three-phase  and  for 

Teaser  PLJr    a  combination  of  three-phase-two- 

phase   transformation   being    con- 

Mam  primary         Mam  secondary          nected  in  a  manner  known  as  the 
FIG.  180.  T-  or  SCOTT  connection. 

T-connection    of  Transformers. 

—This  connection  is  illustrated  in  Fig.  180.  The  connection  is 
used  commonly  in  circuits  with  rotary  converters,  where  a  wire 
may  be  brought  out  from  the  neutral,  h',  and  connected  to  the 
middle  wire  of  a  three-wire  system  on  the  direct-current  side. 
In  this  case  the  direct  current  flowing  in  the  transformer  wind- 


238 


ELECTRICAL  ENGINEERING 


ings  has  no  magnetizing  effect  since  it  flows  in  opposite  direc- 
tion in  the  two  halves  of  the  transformer  windings. 

It  consists  of  a  so-called  "main"  transformer  with  a  tap 
brought  out  at  the  middle  points  of  its  windings  and  a  " teaser" 
transformer  of  0.866  times  as  many  turns,  one  terminal  of  which 
is  connected  to  the  tap,  d,  of  the  main  transformer.  The  three- 
phase  lines  are  brought  to  the  terminals,  a,  b,  c,  which  are  at  the 
three  vertices  of  an  equilateral  triangle.  Thus,  if  the  base,  ac, 
of  the  triangle  has  the  length,  Z,  its  height,  bd,  will  be  0.866. 
The  center  of  this  triangle  will  be  at  a  point,  h,  called  the  neutral. 

Rating  of  T-connected  Transformers. — Three-phase  output 
=  -\/3EI,  where  E  and  /  are  line  voltage  and  current,  respect- 
ively. Rated  output  of  the  two  transformers 

=  El  +  0.866#7  =  1.866#7. 

.".  Ratio  of  the  output  to  the  transformer 

1  73 

rating  is   -T  =  0.925  &  92.5  per  cent.    This 


means  that  for  the  same  values  of  E  and  7, 
FIG.  181.  three  single  transformers  would  need  to  have 

only  92.5  per  cent,  of  the  kva.  rating  which 
the  T-connected  transformers  would  have.  Thus,  the  T-con- 
nection  is  nearly  as  good.  It  may  in  some  cases  be  cheaper,  as 
it  involves  only  two  transformers. 

Two-phase — Three-phase     Transformation. — Let    two-phase 
currents  be  led  to  the  primaries, 
while    three-phases    are    taken 
from     the      secondaries.     Con- 
sidered as  1:1  ratio  of  the  main 
transformers     for     convenience  // 
only,  the  teasers  would  be  in  the' 
ratio  1 : 0.866.    Neglecting  excit- 
ing current  the  two-phase  input  =  2EI,  =  three-phase  output 


=  -7=  7  =  1.167. 


FIG.  182. 


The  rating  of  a  transformer  may  be  taken  as  the  average  of  the 
input  and  output,  and  it  is  therefore, 
Rating  =  %  [2EI  +  1.1QEI  +  (0.866#  X  1.167)] 

=  M  [2#7  +  1.1QEI  +  El]. 

=  El  +  1.08  El  =  2.08  EL 


MULTIPHASE  TRANSFORMER  SYSTEMS        239 

2 

The  so-called  cost  efficiency  is  therefore  ^-7^  =  0.96,  that  is. 

^.Uo 

the  rating  is  96  per  cent,  of  that  of  two  transformers  for  an 
ordinary  two-phase  transformation,  or  for  two  single-phase  trans- 
formers. That  is  nearly  as  good  as  using  three  transformers  for 
the  three-phase  and,  there  being  only  two  transformers,  possible 
economy  is  suggested. 

The  question  arises  as  to  how  it  is  that,  with  such  connections 
the  magnetization  is  uniform.  If  it  is  not  uniform,  there  will 
be  complications  due  to  over  and  under  saturation  in  the  different 
parts  of  the  cores.  Therefore,  the  sum  of  the  magnetomotive 
forces  due  to  the  load  current  in  the  branches  of  the  windings  must 
add  up  to  zero,  that  is,  the  two-phase  load  ampere-turns  in  each 
branch  must  be  balanced  by  the  correspond- 
ing three-phase  ampere-turns.  Let  oa,  ob, 
oc  (Fig.  183),  represent  the  three-phase  AT, 
in  amount  and  direction.  Let  de  represent 
the  two-phase  AT7  in  the  main  transformer. 
To  obtain  the  three-phase  projections  on  the 
two-phase  line,  it  is  necessary  to  take  the  j?IG  183 

components  of  oc  and  ob  on  the  horizontal. 
These  equal  de.     The  components,  however,  are  in  opposite 
directions,  but  due  to  the  fact  that  the  ampere-turns  from  c  to 
6  are  evidently  all  in  the  same  direction,  ob  must  be  projected 
backward  to  ob'.     This  causes  the  vertical  components,  b'd  and 
dc  to  be  in  opposition  and  they  therefore  cancel  each  other. 
The  proof  of  this  by  trigonometrical  relations  is  as  follows: 
m.m.f.  of  primary  =  It  sin  0,  where  t  is  the  number  of  pri- 
mary turns. 

m.m.f.  of  od  =  ~±-  sin  (0  +  210°) 


m.m.f.  of  oe  =  -  sin  (0  -  30°). 

Total  m.m.f.  =  It  sin  0  +  1.16  ~  [sin  (0  +  210°)  +  sin  (0  -  30°)] 

=  It  sin  0  -  0.587*  [sin  (0  +  30°)  +  sin  (0  -  30°)] 
=  It  sin  0  -  0.587*  [sin  0  cos  30°  +  cos  0  sin  30° 

+  sin  0  cos  30°  -  cos  0  sin  30°] 
=  It  sin  0  -  0.587*  (1.73  sin  0)  =  0. 

This  means  that  the  load  current  does  not  increase  the  magneti- 

zation of  the  transformer. 


240  ELECTRICAL  ENGINEERING 

In  the  case  of  the  teaser  transformer,  both  the  primary  and  the 
secondary  are  in  the  same  direction  in  space  and  time,  that  is, 
they  bear  the  same  phase  relation  as  with  single-phase  trans- 
formers. Therefore,  the  ampere-turns  relation  is;  secondary 
A.T.  =  0.866  X  1.16EI  =  El  =  primary  ampere-turns.  There- 
fore as  turns  are  proportional  to  voltage,  the  m.m.fs.  are  equal. 
It  is  always  possible  to  buy  transformers  of  both  10:1  and  9:1 
ratios  from  stock.  For  practical  reasons  9:1  is  used  instead  of 
8.6: 1.  With  these  ratios,  connection  can  be  made  to  nearly  any 
system  in  practical  operation. 

Auto-transformer's    (also   called   compensators). — Auto- trans- 
formers are  transformers  with  only  one  winding.     The  primary 
voltage  is  applied  to  the  coil  terminals;  the  sec- 
ondary voltage  is  obtained  by  connecting  to  taps 
at  any  desired  places  of  the  winding. 

The    general   connections  of  the  single-phase 
auto-transformers  are  as  in  Fig.  184.     Let  /i  and 
J2  be  the  primary  and  secondary  currents,  respec- 
tively. 
Then 

/i  =  current  in  ab. 
/2  —  /i  =  current  in  be. 

The  rating  of  the  section  ab  is  h(Ei  —  Ez).  The  rating  of 
section  be  is  (72  —  Ii)E2.  The  rating  of  the  auto- transformer  is 
the  average  of  the  sum,  or 

rating  =  ^  [/^  -  I±E2  +  I2E2  -  hEz] 

=  1A  [IiEi  +  /2#2  -  2/i^J.  (Ill) 

Neglecting  exciting  current,  as  in  any  transformer,  the  volt- 

"F         I 
age  and  current  ratios  are  -^r  =  7^  or   IiEi  =  I2E2.     Substi- 

•&2          1\ 

tuting  for  I2E2  in  (111),  the  rating  becomes, 

rating  =  %  [2I1E1  -  2/1#2[  =  II  [Ei  -  Et]. 
The  per  cent,  rating  for  a  given  current  is 

Ii(Ei  -  Ei)  =  El  -  Ei 
IjEi  EI 

Thus,  if  E2  =  90  per  cent,  of  Ely 

Per  cent,  rating  = —  -  =  0.1,  or  10  per  cent.     That  is,  it  is 


MULTIPHASE  TRANSFORMER  SYSTEMS       241 

necessary  to  supply  only  10  per  cent,  of  the  rating  of  an  ordinary 
transformer  to  effect  this  transformation  which  is  obviously  a 
great  gain  in  cost  efficiency.  If  the  voltage  is  to  be  reduced  in  the 
ratio  2:1  the  economy  of  using  an  auto-transformer  instead  of  an 
ordinary  transformer  is  not  so  great.  The  saving  is  in  this  case 
about  one-half. 

Problem  83.  —  Show  the  advantage  of  using  auto-transformers  by  plotting 
a  curve  between  per  cent,  rating  of  the  auto-transformers  and  transformation 
ratio. 

Compensators  for  Two-phase  —  Three-phase  Transformation. 

—  In  Fig.  185,  let  the  two-phase  taps  be  cb  and  ef,  and  let  the 
three-phase  taps  be  a,  d,  g,  and  let  Ez  and  /2  be  two-phase  volt- 
age and  current  respectively  and  Es  and  73  be 
corresponding  values  for  three-phase.  Neglecting 


g 


losses,  \/3^3^3  =  2#272,  is  the  power  relation  be-        /  \*\ 
tween  input  and  output.  ,// 

Considering  separate  parts  of  the  windings,  cur-  £-^ 
rent  in  ab  —  73;  voltage  in  ab  =  0.866#3  —  E2,       FIG.  185. 
since  voltage  in   ac  =  0.866#3;   rating  of  ab  = 
(0.866#3  -  #2)73;  current  in  bh  =  72  -  73.     In  this  case  72  > 
73,  E3  being  >  E2.    Voltage  in  bh  =  E2  -  H  0.866  E3  since  he  = 
J^ac;  rating  of  &ft  =  (72  -  73)(#2  -  M  0.866^3);  current  in  he 
=  1 2  —  7s,  since  the  resultant  sum  of  two  equal  currents  120° 
apart  is  numerically  equal  to  one  of  them.     The  three-phase 
current  in  he  is  the  sum  of  the  currents  of  the  phases  hd  and  hg, 
indicated  by  dotted  lines. 

Voltage  in  he  =  l/i  0.866  E3; 

rating  of  he      =  %  0.866  #3  (72  -  /«) ; 

current  in  de    =  73  =  current  infg; 

voltage  of  de    =  H  (#3  -  #2)  =  voltage  of  fg; 

rating  of  de      =  or  (#3  -  #2)  =  rating  of  /gr; 


current  in  ec  =  V(/2  -  h  cos  30°)  2  +  (/8  sin  30")  »  =  current  in 
c/,  that  is,  it  is  72  -  the  component  of  73,  in  phase  with  72  +  j  X 

the  component  of  7s  normal  to  Iz> 

•p 
Voltage  of  ec  =  -~  =  voltage  of  cf. 

Rating  of  ec  =  yV(/2  -  I*  cos  30°)2  +  (/,  sin  30°)2  =  rating 
of  cf. 

16 


242  ELECTRICAL  ENGINEERING 

The  combined  rating,  which  is  one-half  the  sum  of  the  ratings 
of  all  the  parts,  is 

0.933#3/3   ~ 


0.5#2  X  V/22  -  1.73/2/3  +  /32. 

An  examination  of  this  rather  complicated  expression  will  show 
that  the  same  ratio  of  cost  efficiency  holds  with  reference  to  the 
T-connected  transformers  having  primary  and  secondary  wind- 
ings, as  holds  for  single-phase  auto-transformers  compared  with 
ordinary  single-phase  transformers. 

Dissimilar  Transformers  in  Series. — Transformers  may  not  be 
indiscriminately  connected  in  series  with  safety. 

To  connect  two  transformers  of  different  design  but  proper 
rated  voltages  in  series  is  not  always  safe,  since  they  may  not 
take  their  proper  share  of  the  total  voltage.  One  may  even  burn 
out  at  no-load  due  to  excessive  core  loss.  Suppose  that  their 
normal  exciting  currents  are  different  Since  the  same  amount 
of  current  must  flow  through  each  transformer  (as  they  are  in 
series),  this  current  will  be  insufficient  to  give  the  proper  flux  in 
one  of  the  transformers  and  will  be  more  than  necessary  in  the 
other.  Thus  the  voltages  will  not  divide  according  to  the  rating 
and  the  core  loss  will  be  low  in  one  and  excessive  in  the  other. 
Let  the  open  circuit  or  exciting  impedance  of  A,  Fig.  186,  be 

r   +  jx    =  z, 
that  of  £,  A 


~\-  JXi   =  Z\.  B 

The  total  impedance  is  then  Z=z  +  Zi  =  .R-f  jX.       FIG.  186. 
Then  the  exciting  current  of  the  two  transformers  in  series  is 

,    _  CQ  _  impressed  volts 
"  Z  =         R+jX 

The  voltage  across  A,  is 
Voltage  across  B  is 

VB=     e° 


Neglecting  the  power  component,  we  get  as  a  fair  approximation, 


MULTIPHASE  TRANSFORMER  SYSTEMS       243 


snce 


'  approximately, 


A   -• 


_ 

/.I 


Im,A 


where  Im,B>  and  7mrA,  are  the  normal  magnetizing  currents.  Thus, 
the  respective  voltage  drops  across  A  and  B,  when  in  series,  will 
be  approximately  inversely  proportional  to  the  normal  magnetiz- 
ing or  exciting  currents. 

This  assumes  constant  values  of  x  and  rti,  which  would  not  be 
true  if  the  resultant  exciting  current  differed  widely  from  the 
normal  values  of  exciting  current  of  the  two  transformers. 

If  x  is  nearly  equal  to  x\9  the  above  ratios  would  hold.  Where 
one  transformer,  however,  is  saturated,  its  reactance  is  greatly 
diminished,  which  allows  a  greater  current  to  flow  in  the  circuit  but 
tends  to  equalize  the  voltages. 

Dissimilar  Transformers  in  Parallel.  —  This  is  the  usual  mode 
of  connection  and  it  offers  no  difficulty  due  to  unequal  exciting 
currents.  The  question,  here,  is  one  of  proper  division  of  the 
load.  In  giving  orders  for  additional  equipment,  it  is  customary 
to  specify  what  the  percentage  reactance  of  the  new  transformers 
shall  be.  With  equal,  or  proportional,  reactances  there  results 
a  proper  division  of  the  load.  Consider  the  parallel  connection 
as  shown  in  Fig.  187. 

The  two  load  currents  are: 

I  A  =  i  +  Ji', 
IB  =  i\  +  fi'i- 

The  total  load  current, 

I  =  IA  +  IB. 

'  ' 

Let 

ZA  =  r  +  jx  and  ZB  =  TI  + 


FIG.  187. 


be  the  impedances  of  A  and  B  respectively. 

Then,  since  the  terminal  voltages  are  the  same  on  each,  the 
voltage  drops  in  the  transformers  are  equal,  and  are: 

(i  +  ji')(r  +  jx)  =  (ii  +  ji'i)(ri  +  jxi). 


244  ELECTRICAL  ENGINEERING 

Multiplying  out, 

ir  +  jix  +  ji'r  -  i'x  =  itfi  +  ji&i  +  ji'tfi  -  i'&i. 
Here,  the  real  components  must  be  equal  and  the  imaginary 
components  must  be  equal. 

.".  ir  -  i'x  =  itfi  -  i'&i, 
and 

ix  +  i'r  =  iiXi  -j-  zVi. 

Neglecting  resistances, 


and 

ix  = 

whence 

•/  • 

i>  X\        i>  3/1 

z'i  "~  a; '   ii  "~  # 

Thus,  the  load  is  divided  in  inverse  proportion  to  the  reactances. 

The  best  method  of  connection  is,  as  in  Fig.  188.     The  student 

is  advised  to  explain  why  this  is  so.    '    A  ^c 

ft  P  i 


FIG.  188. 

Three-phase  Connection  of  Dissimilar  Transformers.  —  If 
the  three  transformers  are  connected  Y—  Y  there  will  not  be 
symmetrical  distribution  of  voltage.  Consider  the  neutral 
point,  0,  Fig.  189,  with  reference  to  the  transformers  A  and  B. 
With  line  voltage  impressed  on  AB,  the  potential  at  0  may  have 
any  intermediate  value,  just  as  with  two  single-phase  trans- 
formers in  series,  depending  on  the  relative  open  circuit  im- 
pedances of  the  two  transformers. 

The  point  of  junction  of  the  three  transformers  may,  for  in- 
stance, be  displaced  to  0'.  The  secondary  Y-voltages  would 
have  a  similar  relationship  to  each  other.  Dissimilarity  may 
consist  merely  in  variation  in  the  iron  of  two  supposedly  similar 
transformers. 

If  the  secondaries  are  connected  in  A,  the  primaries  being 
Y-connected  this  difficulty  of  unbalanced  potentials  is  eliminated. 

The  induced  voltage  in  each  secondary  will,  of  course,  be 


MULTIPHASE  TRANSFORMER  SYSTEMS        245 


proportional  to  that  of  its  primary,  giving  the  closed  A,  ABC, 
Fig.  190,  when  the  primary  circuit  is  balanced. 

With  unbalanced  condition,  if  the  A  is  left  open  at  B,  the  vol- 
tage vectors  will  not  make  a  closed  figure,  but  as  shown  by  the 
dotted  lines,  will  leave  an  opening  between  B'  and  B".  If  the 
A  is  then  closed,  the  voltage  B'B"  will  act  in  the  A  circuit,  sending 

a  local  current  which  will  increase  the 
magnetization  of  the  transformer  whose 
flux  is  below  normal  and  decrease  that 
of  the  transformer  whose  flux  is  above 


B' 


FIG.  190. 


FIG.  191. 


normal.  Thus  the  magnetization  is  brought  back  to  normal 
value,  the  local  current  in  the  A  serving  to  anchor  the  neutral 
point  of  the  Y. 

Three-phase  transformer  systems  may  be  extended  in  a  variety 
of  ways  to  cover  cases  where  it  is  desirable  to  use  six  phases. 
This  practice  finds  application  especially  with  rotary,  or  syn- 
chronous, converters,  and  it  will  be  discussed  more  fully  under 
that  heading.  Such  combinations  of  transformers  as  permit 
symmetrical  grouping  of  voltages  are  illustrated  by  the  double  A, 
double  T,  or  double  Y  shown  in  Fig.  191. 


i   2   3 


(a) 


CHAPTER  XXXIII 
ALTERNATORS 

Fundamentally,  direct-current  and  alternating-current  genera- 
tors are  alike.  An  alternator  becomes  a  direct-current  generator 
by  adding  a  commutator.  The  essential  principles  of  both 
machines  have  been  developed  in  Chaps.  VI  and  VII. 

In  Fig.  192,  a,  is  represented  a  simple  alternator  with  a  two-pole 
field  core  magnetized  with  direct  current  from  some  independent 

source,  and  an  armature  with 
a  single  coil.  As  this  armature 
revolves  there  is  generated  in  it 
an  e.m.f.  which  follows  closely 
a  sine  wave  of  time  values.  It 
is  apparent  that  the  space  on 
the  armature  periphery  is  not 
all  utilized,  and  that  another 
coil  could  be  put  on  in  space 
quadrature  to  the  first.  In 
such  a  case,  two  similar  e.m.f.  waves  would  bfi  produced  but  in 
time  quadrature  with  each  other,  or  at  90°  time-phase  displace- 
ment, that  is,  one  wave  would  reach  its  maximum  one-quarter 
of  a  period  later  than  the  other  (Fig.  192,  6). 

Such  an  alternator  is  called  a  two-phase,  or,  sometimes,  a 
quarter-phase  machine. 

On  the  same  principle,  an  armature  may  be  supplied  with  three 
coils,  or  groups  of  coils,  spaced  120°  apart,  each  group  giving  its 
separate  e.m.f.  wave  (Fig.  192,  c). 

In  this  way,  any  number  of  coils  or  groups  of  coils  may  be 
wound  on  an  armature,  giving  any  desired  number  of  phases. 
In  practice,  however,  the  majority  of  alternators  are  three- 
phase,  and  very  seldom  is  one  built  for  a  greater  number  of 
phases  than  three. 

The  voltages  generated  in  the  various  phase  windings  may  be 
conveniently  shown  in  their  proper  relations  by  vectors. 

If  in  the  two-phase  case,  the  ends  1'  and  2'  (Fig.  193,  6),  are 

246 


(6) 
FIG.  192. 


ALTERNATORS  247 

joined  together,  the  voltages  of  the  two  coils  will  be  added 
vectorially,  so  that  a  voltmeter  placed  across  the  terminals,  1,  2, 
would  read  \/2  times  the  voltage  of  either  coil  taken  separately, 
since  the  two  voltages  are  in  time  quadrature.  Likewise  by 
connecting  1  and  2',  the  joint  reading  across  1'  and  2  will  obvi- 
ously also  be  \/2  times  the  voltage  of  one  phase. 

With  a  three-phase  machine  it  is  not  quite  so  apparent  that  the 
voltage  between  the  three  collector  rings  is  \/3  times  the  voltage 
generated  in  one  phase.  At  first  sight  it  might  be  expected  that 
the  resultant  voltage  should  be  the  same  as  that  generated  in 
each  phase  since  the  voltages  are  120°  apart. 


C 

FIG.  194. 

Let,  in  Fig.  194,  OA,  the  voltage  of  phase  A,  be  represented  by 

eA  =  Em  sin  at. 
Then  OB,  the  voltage  of  phase  B,  is  evidently 

eB  =  Em  sin  (at  +  120), 
and 

ec  =  #msin  (orf  +  240). 

The  voltage  between  collector  rings  A  and  B  is  thus 

SA  —  eB  =  Em  [sin  at  —  sin  (co£  +  120)], 
which,  by  simple  trigonometric  transformation  becomes, 

€A  _  €B  =  V3#m  sin  (w<  -  30°). 

Thus  the  numerical  value  of  the  potential  difference  between  the 
collector  rings  is  \/3  times  as  great  as  the  voltage  generated  in 
each  phase  and  the  resultant  voltage  is  displaced  30°  from  the 
voltage  generated  in  phase  OA. 

Problem  84. — Prove  that  the  voltage,  between  B  and  C  is: 

\/3Em  sin  (wt  +  90), 
and  that  the  voltage  between  C  and  A  is: 

\/3Em  sin  (ut  +  210). 

It  is  seen,  thus,  that  the  voltages  between  the  collector  rings  are  also 
120°  apart. 


248 


ELECTRICAL  ENGINEERING 


It  is  interesting  to  note  here  that  a  single-phase  machine  might 
be  treated  as  a  two-phase  machine  in  which  the  two  phases  are 
180°  apart  as  is  shown  in  Fig.  195. 

Let  the  voltage  generated  in  OA   be  Em.sin  co£.     Then  that 
^  generated  in  OB  is  Em  sin  (coZ  +  180).     Thus 

~B~~       o       ~~5  the  difference  of  potential  between  the  collector 
FIG.  195.         rings  at  A  and  B  is 


(sin  (at  —  sin  («'  +  180)  =  2Em  sin  ut. 


The  resultant  potential  difference  is  twice  the  voltage  generated 
in  each  phase,  as  should,  of  course,  be  the  case. 

This  fact  could  have  been  developed  also  geometrically. 

To  find  the  potential  difference  between  A  and  B}  Fig.  194,  we 
should  subtract  OB  from  OA  as  shown  in  Fig.  196. 

It  is  not  necessary  that  the  windings  shall  consist  of  separate 
coils.  A  closed  ring  winding,  or  Gramme  ring,  may  be  tapped  at 
symmetrical  points  and  these  connected  to  slip  rings,  as  in  Fig. 
197.  Thus,  if  a  voltmeter  is  connected  across  the  slip  rings  (1,1), 
the  voltage  of  one  phase  is  read.  If  connected  across  rings  (2,  2), 
the  same  value  of  voltage  will  be  indicated,  but  it  is  evident  that 
the  phase  of  this  e.m.f.  is  displaced  by  90  time  degrees  from  that 
of  the  first. 


FIG.  196. 


FIG.  197. 


With  the  three-phase  connection  taps  are  brought  out  at 
points  120  space  degrees  apart  and  led  to  slip  rings.  A  volt- 
meter, connected  across  any  two  rings,  will  read  the  voltage 
of  one  coil,  say  coil  a,  Fig.  197.  But  this  must  also  be  the  sum  of 
the  voltages  of  the  other  two  coils,  since  any  one  coil  is  in  parallel 
with  the  other  two  coils  with  respect  to  the  external  circuit. 
The  voltages  in  this  case  form  a  closed,  so-called  delta,  A,  and  it 
is  evident  that  the  phase  voltage  and  line  voltage,  or  voltage 
between  the  collector  rings,  are  equal. 

In  these  diagrams  only  three  collector  rings  and  three  lines  are 
shown.  Yet  in  the  discussion  it  has  been  assumed  that  one  side 


ALTERNATORS  249 

of  each  winding  is  connected  to  a  common  point.  It  would  seem, 
therefore,  that  at  least  four  collector  rings  and  lines  might  be 
necessary  to  form  a  complete  system,  in  other  words,  that  even  a 
balanced  three-phase  system  would  involve  four  wires  as  is  shown 
in  Fig.  199.  It  is  evident  that  if,  with  a  balanced  system,  the 
current  in  an  ammeter  placed  at  N  is  always  zero,  then  no  return 
or  fourth  wire  is  necessary. 


FIG.  198.  FIG.  199. 

Let  the  current  in  phase  A  be 

ia  =  Im  (sin  a>t)  and  the  current  in  phases 
B  and  C  be  ib  =  Im  sin  (orf  +  120)  and  ic  =  Im  sin  (at  +  240). 
The  current  in  N  is  then 

in  =  ia  +  ib  +  ic  =  Im  sin  0=0. 

Problem  86. — Prove  that  no  circulatory  current  of  fundamental  frequency 
flows  in  the  delta-connected  generator. 

Since  with  the  Y-connected  generator  the  transmission  lines  really  form 
extensions  of  the  windings,  it  is  evident  that  whatever  current  flows  in  the 
line  also  flows  in  each  winding. 

With  the  delta-connected  generator  this  is  not  so,  because  the  line  current 
is  the  vector  sum  of  the  currents  in  the  adjacent  phases,  as  is  shown  in  Fig. 
198.  The  current  in  phase  1-2  may  be  considered  the  zero  vector.  Thus 
the  currents  in  the  phases  are: 

in  =  Im  sin  (at, 
i*  =  Im  sin  (at  +  120), 
iai  =  Im  sin  (tat  +  240). 
Then,  since  the  sum  of  the  currents  flowing  to  a  point  is  zero, 

it  +  las  -  in  =  0, 
or  it  =  iiz  —  iza  —  Im  sin  wt  —  Im  sin  (wt  +  120) 

=  Vzlm  sin  (cat  -  30). 

The  line  current  is  thus  \/3  times  as  large  as  the  current  in  the  individual 
phases. 

Referring  to  Fig.  198,  it  is  evident  that 

is  +  in  —  *32  =  0  and  i\  +  in  —  isi  =  0. 

Problem  86.— Prove  that  the  currents  in  lines  1  and  3  are,  respectively, 
sin  (tat  +  90)  and  \/3/m  sin  (tat  +  210). 

The  power  given  by  a  three-phase  alternator  is 
P  =  \/3EI  cos  a, 


250  ELECTRICAL  ENGINEERING 

whether  the  alternator  is  connected  Y  or  A,  where  /  is  the  effec- 
tive value  of  the  line  current  and  E  the  effective  value  of  the 
voltage  between  the  lines,  and  a  is  the  angle  of  lead  or  lag  of  the 
phase  current  in  reference  to  the  phase  voltage,  that  is,  cos  a  is 
the  power  factor. 

To  prove  this,  consider  a  Y-connected  generator. 

Since  I  is  the  line  current,  it  is  also  the  current  in  each  winding. 

Since  E  is  the  line  voltage,  the  voltage  of  each  of  the  three 

E 

phases  of  the  generator  is  —=?     Thus  the  power  given  by  each  of 

V3 

•pr 

the  three  phases  of  the  generator  is  — =  cos  a,  and  the  total 

V3 

El 

power,  3  -—=•  cos  a,  =  \/?>EI  cos  a. 
v  3 

Problem  87. — Prove  that  this  also  applies  in  the  case  of  a  delta-connected 
generator. 

Voltage  to  Neutral. — In  Y-connected  alternators  the  neutral 
point  is  the  center  of  the  Y.  On  a  three-phase  distribution 
system  it  is  often  advantageous  to  run  a  fourth  wire  from  the 
neutral. 

The  voltage  between  any  of  the  other  wires  and  the  neutral 
is  the  phase  voltage,  and  is  equal  to  the  line  voltage  divided  by 

Vs. 

In  a  A-connected  alternator  there  is  no  actual  neutral  point. 
However,  for  purposes  of  calculation,  a  neutral  point  is  imagined 
at  the  center  of  the  delta,  and  the  voltage  to  neutral 
is  then  the  phase  voltage  divided  by  \/3,  or,  since 
the  phase  voltage  and  the  line  voltage  are  the  same, 
it  is  equal  to  the  line  voltage  divided  by  \/3  as  with 
Y-connected  alternator.     The  voltage  to  neutral  is 
thus  independent  of  the  manner  of  connecting  the  alternator 
windings. 

Rating  of  Alternators. — As  with  transformers,  alternators  are 
rated  in  kilovolt-amperes,  not  in  kilowatts.  This  is  because 
the  permissible  output  of  an  alternator  depends  on  the  current  in 
its  windings,  regardless  of  the  phase  relation  between  the  current 
and  the  voltage. 

The  nominal  rating  of  an  alternator  may  be  designated  as,  for 
example,  A.T.B.  12-400-600-2300,  where  A  signifies  alternator, 
T  signifies  three-phase,  (S  is  for  one  or  single-phase),  B  signifies  a 


ALTERNATORS  251 

revolving  field.     If  the  armature  is  the  revolving  part,  the  third 
letter  is  omitted. 

12  signifies  the  number  of  poles. 
400  signifies  the  rating  in  k.v.a. 
600  signifies  the  speed  in  r.p.m. 
2300  signifies  the  rated  voltage. 

Sometimes  a  subscript  is  added  to  the  second  letter;  thus, 
A  TZ  signifies  a  three-phase  revolving  armature  alternator  having 
two  slots  per  pole  per  phase  on  the  armature. 

If  the  above  alternator  is  Y-connected,  the  phase  voltage,  or 

2300 
voltage  to  neutral,  is  —  -1=  —  1330  volts. 

400  k.v.a.          400,000 
The  line  and  phase  current  is  23QQ   =  3^1330  =  10° 


amp. 

If  delta-connected,  the  voltage  to  the  imaginary  neutral  is 

likewise  1330. 

100 

The  line  current  is  also  100,  but  the  phase  current  is  —/=  = 

v  3 

57.7  amp. 


CHAPTER  XXXIV 
ARMATURE  REACTION 

The  so-called  armature  reaction  of  a  machine  is  a  measure  of  the 
m.m.f.  of  the  armature.  It  is  thus  expressed  in  so  many  ampere- 
turns,  either  on  the  whole  circumference  of  the  armature  or, 
more  often,  the  m.m.f.  on  one  pole  of  the  armature.  This  latter 
convention  will  be  used  in  this  book. 

As  will  be  seen,  the  m.m.f.  of  the  armature  current  sometimes 
acts  against  the  m.m.f.  of  the  field  excitation,  sometimes  it 
assists  it,  and  often  its  effect  is  only  to  shift  the  flux. 

In  Fig.  201  the  coil  (1,  1)  is  in  the  position  of  zero,  or  mini- 
mum, e.m.f.,  assuming  the  flux  to  be  symmetrical  in  the  field 
system,  or  due  to  the  field  ampere-turns  alone.  The  coil  (2,  2) 
is  in  the  position  of  maximum  e.m.f.  This  condition  may  be 
assumed  to  hold  for  no-load.  The  coil  (3.,  3)  is  in  an  interme- 
diate position.  The  current  may  or  may  not  be  in  time-phase 


FIG.  201. 


FIG.  202. 


with  the  e.m.f.,  but  whatever  its  time-phase  relation  may  be, 
in  spa6e,  it  is  evident  from  Fig.  201  that  the  m.m.f.  of  the  coil 
is  at  right  angles  to  the  surface  of  the  coil,  and  therefore  at  right 
angles  to  the  line  which  represents  the  position  of  the  coil. 

In  Fig.  202  let  the  armature  current  lag  behind  the  e.m.f.  Its 
m.m.f.  is  seen  to  be  largely  in  opposition  to  that  of  the  field  which 
causes  the  main  flux,  and  this  opposition  increases  the  greater  the 
lag  and  becomes  complete  at  90°  lag.  Similarly  a  leading  current 

252 


ARMATURE  REACTION 


253 


umru 

•S" 

FIG.  203. 


increases  the  flux.  The  current,  i,  may  be  divided  into  two. 
components,  one  of  which,  i",  is  entirely  wattless  and  exactly 
opposes  the  field  flux,  and  the  other,  i1 ',  the  watt  component  in 
phase  with  e{,  which  merely  distorts  the  field.  The  effect  of 
current  in  the  armature  is  to  weaken  the  resultant  flux  and  to 
displace  its  maximum  position,  if  the  current  lags,  as  shown  in 
Fig.  203.  The  weakening  is  due  to  the  wattless 
component,  the  displacement  or  distortion  to  the 
power  component. 

Thus  the  trailing  pole-tip  may  even  become 
saturated,  while  the  leading  pole-tip  is  robbed  of 
a  large  part  of  its  flux.  The  position  of  the  coil 
for  maximum  induced  e.m.f.  is  shifted  ahead,  with 
lagging  current,  and  behind,  with  leading  current. 
These  relationships  are  shown  in  Fig.  204,  where 
the  induced  e.m.f.,  eiy  is  taken  as  the  zero  vector. 
6i  is  at  right  angles  to  the  resultant  flux,  fa,  and  lags  behind  it. 
The  armature  current,  /,  is  taken  at  any  angle  and  produces  a 
flux  (f>a  in  time-phase  with  it.  This  flux  vectorially  subtracted 
from  fa,  gives  <£/,  the  field  flux.  In  phase  and  90°  in  time 
ahead  of  the  current  are  respectively,  Ir  and  Ix,  the  e.m.f. 
consumed  by  the  armature  resistance  and  that  consumed  by 
the  armature  reactance.  These  com- 
bine to  make  Iz,  the  impedance  drop, 
which,  subtracted  from  et,  gives  e  the 
terminal  voltage.  The  phase  angle  of 
the  load  is  then  0. 

In  constructing  the  diagrams  there  is 
some  advantage  in  using  ampere-turns 
instead  of  fluxes,  since  then  no  compli- 
cations arise  from  variable  magnetic 
reluctances.  This  has  been  done  in 
Fig.  207,  Chap.  XXXV. 

At  no-load,  I  =  0,  et  =  e  and  fy  =  fa. 
In  Fig.  204  the  angle  y  is  the  angular 
space  displacement  of  the  armature  with  respect  to  the  field 
poles,  due  to  the  load,  that  is,  the  angle  between  <f>f  and  fa, 
or,  7  =  /?  —  90°,  where  0  is  the  angle  between  et  and  <£/.  a  is 
the  angle  between  et  and  e.  Considered  on  the  basis  of  experi- 
mental data,  e,  /,  r  and  x  are  known  and  e  is  chosen  as  the  zero 
vector. 


I 


FIG.  204. 


254  ELECTRICAL  ENGINEERING 

Then  the  induced  e.m.f., 
Ei  =  e  +  IZ  =  e  +  (i  +  jii)(r  +  jx) 

—  e  +  ir  +  jix  +  j't'ir  —  iiS 

=  (e  +  ir  -  iix)  +  j(ix  +  i»  =  a  +  j6  (112) 
where  I  =  i  +  jii  for  leading  current,  and  ii  is  negative  for 
lagging  current.  From  the  saturation  (magnetization)  curve  of 
the  generator  the  number  of  ampere-turns  needed  to  produce  this 
voltage  is  found.  Let  Fr  —  the  resultant  m.m.f.  =  C  (a  +  jb) 
or,  rather,  Fr  —  jC(a  +  jb)  because,  in  space,  as  has  been  shown, 
the  m.m.f.  is  rotated  90°  with  respect  to  Ei. 
Then, 

Fr  =  C(-  b+ja).    _ 

But  the  resultant  m.m.f.  is  the  vector  sum  of  the  field  and 
armature  m.m.fs.  That  is, 

Fr  =  Ff  +  Fa  =  Ff  -f  m  (i  +  jii)  (113) 

Thus,  C  is  the  proportionality  factor  between  the  resultant 
field  ampere-turns  and  the  volts,  and  m  is  that  between  the 
armature  ampere-turns  and  the  current. 

Examples— If  Et  =  2500  volts   and    Fr  =  3000  amp.-turns, 
c  _  3000  _ 
C  "2500" 

,.                V2  X  1.5  X  It       010. 
In  a  three-phase  machine,  m  =  —     — j =  2.12£,  where 

i  =  turns  per  phase  on  the  armature.  (This  factor  is  discussed 
more  fully  later.)  The  quantity,  \2,  enters  in  order  to  derive 
the  maximum  value  of  the  ampere-turns  from 
the  effective  value;  1.5  comes  from  the  fact 
that  the  resultant  field  of  a  three-phase  system, 
as  an  armature,  or  induction  motor  field,  is  1.5 
times  that  of  a  single-phase  system.  This  is 
shown  as  follows:  Consider  the  components 
of  flux  along  the  two  axes  (Fig.  205).  The  three 
x-components  in  space  are  H,  H  cos  120°,  H  cos  240°.  The 
components  along  the  i/-axis  in  space,  are  0,  H  sin  120°,  H  sin 
240°.  The  components  of  all  the  phases  in  time  are  H  cos  8, 
H  cos  (0  +  120°)  and  H  cos  (0  +  240°). 

Hence  the  sum  of  components  along  the  z-axis  in  time  and 
space  is  H  cos  0  +  H  cos  (0  -f  120°)  cos  120°  +  H  cos  (0  +  240°) 
cos  240°  =  1.5#  cos  B. 


ARMATURE  REACTION         .  255 

The  sum  of  components  along  the  2/-axis  in  time  and  space,  is 
0  +  H  cos  (0  +  120°)  sin  120°  +  H  cos  (0  +  240°)  sin  240°  = 
1.5tf  sin  0. 

Transposing  (113),  the  field  m.m.f.  is 
Ff  =  Fr  -  m(i  +  jii) 
=  C(-b+ja)  -  m(i 
=  —  bC  +  jaC  —  mi  — 
=  -  60  -  mi  +  j(aC  -  mi^)  (114) 

(With  lagging  current,  ii  is  negative.) 

Numerically,  Ff  =  \(  -  bC  -  mi)2  +  (aC  -  mt\)2        (115) 

Problem  88. — In  a  certain  alternator  let  the  reactance  drop  be  10  per  cent., 
the  resistance  drop  2  per  cent,  and  the  armature  reaction  equal  to  no  load 
one-half  the  field  ampere-turns.  How  many  ampere-turns  are  required  in 
the  field  winding  when  the  alternator  is  carrying  full-load  current  at  80 
per  cent,  power  factor? 

Since 'the  drops  are  given  in  percentage,  e  will  be  taken  =  1,  and  7  =  1. 
Then  at  0.8  P.F., 

i  =  0.8,  ii  =  0.6, 
and  a  =  e  -\-  ir  —  i\x 

=  1  +  0.016  +  0.06  =  1.076 
and  b  =  ix  +  i&  =  0.8  -  0.12  =  0.68. 

On  the  percentage  bases,  also,  let 

C  =  1. 
Then  m  =  0.5. 

The  ampere-turns  required  for  the  field  will  then  be 

Ff  =  V(-  0.68  -  0.4)2  +  (1.076  +  0.3)2  =  1.45. 

For  non-inductive  load,  ii  =  0  and  i  =  1.     Then 
Ei    =  Va*  +  b*  =  1.023, 
and  Ff  =  1.186. 

As  a  continuation  and  amplification  of  this  problem,  consider  the  follow- 
ing: 

Problem  89. — A  three-phase  generator  has  2  per  cent,  resistance  and  10  per 
cent,  reactance.  Its  armature  reaction  is  one-half  the  no-load  field  ampere- 
turns.  The  magnetic  reluctance  is  uniform  all  around  the  periphery  and  the 
saturation  curve  is  a  straight  line  through  the  origin,  at  the  point  of 
operation. 

Plot  the  field  excitation  (a)  against  armature  current,  with  variable  non- 
inductive  load,  up  to  high  overloads;  (b)  at  full-load  current,  but  with  vari- 
able power  factor — leading  and  lagging;  (c)  with  full-load  power  output  and 
variable  power  factor;  (d)  same  as  (a)  but  at  20  per  cent,  higher  voltage;  (e) 
same  as  (6)  but  at  20  per  cent,  higher  voltage;  (/)  same  as  (c)  but  at  20  per 
cent,  higher  voltage;  (g)  same  as  (a)  but  at  80  per  cent,  of  rated  voltage;  (/O 
same  as  (b)  but  at  80  per  cent,  of  rated  voltage;  (i)  same  as  (c)  but  at  80  per 
cent,  of  rated  voltage. 


256 


ELECTRICAL  ENGINEERING 


By  equation  (115) 


Ff  =  V  (-  bC  -  mi)2  +  (aC  -  mil)2 
where 

C  =  1,  m  =  0.5 
r  =  0.02,  x  =  0.1 
o  =  e  +  ir  —  ii#, 
5  =  ix  -f-  iir 

(o)  e  =  1,  i  =  variable,  ii  =  0. 

f  (l+0.02i)2~  = 


-  0.6i)2  + 

=  \/0.36i2  +  1  +  0.04*  +  0.0004t2  =  \/l  +  0.04i  +  0.3604^ 
Tabulating: 


i 

0  0 

0  25 

0  5 

0  75 

1  0 

1  25 

1  5 

2  0 

0.04i  .... 
i»  

0.0 
0.0 

0.01 
0.0625 

0.02 
0.25 

0.03 
0.5625 

0.04 
1.0 

0.05 
1.56 

0.06 
2.25 

0.08 
4.0 

0.3604i*  . 

£"'  

0.0 
1.0 

1.0 

0.0225 
1.0325 
1.014 

0.09 
1.11 
1.052 

0.2027 
1.2327 
1.109 

0.3604 
1.4004 
1.182 

0.562 
1.612 
1.269 

0.811 
1.871 
1.366 

1.442 
2.522 
1.587 

(d) 


1.2;  Ff  =  \/1.44  +  QJ048i  +  0.3604*2 


048f  

0.0 

0.012 

0.024 

0.036 

0.048 

0.06 

0.072 

0.096 

(F/)»  

1.44 

1.4745 

1.554 

1.6787 

1.8484 

2.062 

2.323 

2.978 

Ff 

1  2 

1  211 

1  244 

1  293 

1  359 

1  433 

1  522 

1  722 

(g)  e  -  0.8;  Ff  =  -y/0.64  +  0.032i  +  0.3604i2 


0.032i... 

0.0 

0.008 

0.016 

0.024 

0.032 

0.04 

0.048 

0.064 

(f/)i  

0.64 

0.6705 

0.746 

0.8667 

1.0324 

1.242 

1.499 

2.146 

F/  

0.8 

0.819 

0.863 

0.93 

1.014 

1.115 

1.222 

1.463 

(b)  1  =  1,  P.F.  =  ~  =  variable,  e 


-  0.6i  -  0.02ii)2 


+  0.02i  - 


P.F  
» 

0.0 

o  o 

0.25 
0  25 

0.5 
0  5 

0.75 
0  75 

1.0 
1  0 

0.75 
0  75 

0.5 
0  5 

0.25 
0  25 

0.0 
0  0 

ti  

1  0 

0  968 

0  866 

0  661 

0  0 

0  661 

0  866 

0  968 

1  0 

0.6i  
0.02n  .  .  . 
*  
«*  

0.0 
0.02 
-0.02 
0  0004 

0.15 
0.01936 
-0.16936 
0  0288 

0.3 
0.01732 
-0.3173 
0  1008 

0.45 
0.01322 
-0.4632 
0  215 

0.6 
0.0 
-0.6 
0  36 

0.45 
-0.01322 
-0.4368 
0  191 

0.3 
-0.01732 
-0.2827 
0  08 

0.15 
-0.01936 
-0.13064 
0  017 

0.0 
-0.02 
0.02 
0  0004 

0.02i  .... 
-0.6n.. 
t  
1* 

0.0 
-0.6 
0.4 
0  16 

0.005 
-0.58 
0.425 
0  181 

0.01 
-0.52 
0.49 
0  24 

0.015 
-0.3965 
0.6185 

0000 

0.02 
0.0 
1.02 

0.015 
0.3965 
1.4115 

2A-\ 

0.01 
0.52 
1.53 
2  sue 

0.005 
0.58 
1.585 

0.0 
0.6 
1.6 

2C£ 

«*  +  <2... 
Ff  

0.1604 
0.40 

0.2098 
0.4575 

0.3408 
0.584 

0.598 
0.772 

1.4 

1.182 

2.191 
1.48 

2.43 
1.56 

2.532 
1.59 

2.5604 
1.6 

(e)  e  -  1.2,  Ff  =  \/(-  0.6i  -  0.02ii)2  +  (1.2  +  0.02i  -  0.6ii)2  -  Vs*  +  <i2 


tl.  . 

0  6 

0  625 

OftQ 

0010C 

Inn 

tl*  

0  36 

0  390 

0  476 

0  67 

1400 

2fi 

3    A 

30 

3     OK 

«*  +  1*... 

Ff.  . 

0.3604 
0  6 

0.4188 
0  646 

0.5768 

07EC 

0.885 

004. 

1.852 

IOfi 

2.791 

3.08 

3.217 

3.2504 

(A) 

e  -  0.8, 

Ff  =  V 

-  0.6i  - 

0.02*02 

+  (0.8  -f 

0.02*  - 

0.6ti)2  = 

Vs2  +  t 

22 

fl 

0  2 

0  225 

Ff...2.  '.'. 

0.04 
O.C404 
0  2 

0.0508 
0.0796 
0  282 

0.084 
0.1848 
0  43 

0.176 
0.391 

Ococ 

0.673 
1.033 

1.475 
1.666 

1.77 
1.85 

1.92 
1.937 

1.96 
1.9604 

(c)  »-l,  P.F. 


=  variable,  e  =  1. 


ARMATURE  REACTION 


257 


Ff  =       (-  0.6i  -  0.02ii)2 


0.02i  -  6ii)2  = 


P.F  

0.25 

3.  87 

0.5 

1.73 

0.75 
0.883 

1.0 
0.0 

0.75 
—0.883 

0.5 

-1  73 

0.25 
—  3  87 

0.02n.... 

0.0774 
-0  6774 

0.0346 
-0.6346 

0.01766 
-0.6177 

0.0 
-0  6 

-0.01766 
—  0  5823 

-0.0346 
—  0  5654 

-0.0774 
—  0  5226 

s2  
-O.Gii.  . 
t  

t2  . 

0.46 
-2.322 
-1.302 
1  7 

0.402 
-1.04 
-0.02 
0  0004 

0.382 
-0.53 
0.49 
0  24 

0.36 
0.0 
1.02 
1  04 

0.34 
0.53 
1.55 
2  4 

0.32 
1.04 
2.06 
4  25 

0.274 
2.322 
3.342 
11  2 

s2  +  *2.  . 
Ff  

2.16 
1  47 

0.4024 
0  634 

0.622 
0  788 

1.4 

1  181 

2.74 
1  65 

4.57 
2  14 

11.474 
3  38 

(/)  e  =  1.2,  Ff  =  \/s2  +  (1.2  +  0.02t  -  0.6ii)z  = 


1  1    

-1.102 

0  18 

0  69 

1  22 

1  75 

2  26 

3  542 

fl2           

1  22 

0  0325 

0  476 

1  49 

3  07 

5  11 

12  6 

S2   +  *12.. 

Ff  

1.68 
1  295 

0.4345 
0  659 

0.858 
0  925 

1.85 
1  36 

3.41 
1  846 

5.43 
2  33 

12.874 
3  59 

(i)  e  =  0.8,  Ff  - 


+  (0.8  +  0.02i  -  0.6ii)2  - 


tz  ... 

—  1  502 

—0  22 

0  29 

0  82 

1  35 

1  86 

3  142 

fz2  

S2  +  <22.. 

Ff  

2.26 
2.72 
1.65 

0.0485 
0.4505 
0.671 

0.0841 
0.4661 
0.682 

0.673 
1.033 
1.015 

1.82 
2.16 
1.47 

3.45 
3.77 
1.94 

9.9 
10.174 
3.182 

Curves  showing  the  variations  brought  out  by  this  problem  are  given  in 
Fig.  206. 

2.4 


0.25 


0.50 


0.25        0.50 
Leading 


0.75         1.0         1.25 

Current 
0.75         1.0         0.75 

Power  Factor 

FlQ.  206. 


1.50 


0.50        0.25 

Lagging 


2.00 

0 


17 


CHAPTER  XXXV 

CHARACTERISTICS  OF  ALTERNATORS  WITH  DEFINITE 

POLES 

In  the  preceding  chapter  it  has  been  assumed  that  the  magnetic 
reluctance  is  uniform  in  the  direction  of  the  main  field  magneto- 
motive force  as  well  as  in  the  transverse  direction  along  the 
armature  surface.  This  condition  exists  practically  in  machines 
with  distributed  field  structures  as  in  induction  generators, 
and,  to  a  very  fair  degree,  turbo-generators. 

In  engine-driven  generators  the  condition  of  uniform  magnetic 
reluctance  rarely  exists,  since  such  machines  are  usually  built 
with  definite  pole  structures.  In  this  type  which  includes  the 
majority  of  machines,  the  magnetic  reluctance  in  the  direction 
of  the  field  poles  is  almost  constant  for  all  m.m.fs.,  and  therefore 
the  flux  is  proportional  to  the  m.m.f.  In  the  direction  along  the 
surface  of  the  armature  the  shift  of  flux  is  by  no  means  pro- 
portional to  the  m.m.f.  but  is  largely  limited1  by  the  mechanical 
construction,  that  is,  the  width  of  the  poles,  and  it  is  always  less 
than  proportional  to  the  m.m.f. 

Similarly,  the  self-induction  of  an  armature  coil  is  greater 
when  it  is  immediately  under  the  poles  than  when  it  is  midway 
between  them.  Thus,  considering  that  the  armature  current  is 
made  up  of  two  components,  one,  the  power  component  which  is 
maximum  when  the  coil  is  under  the  pole,  and  the  other,  the 
wattless  component,  which  is  maximum  when  the  coil  is  midway 
between  the  poles,  it  might  be  assumed  that  the  reactance  is 
greater  for  the  power  component  than  for  the  wattless  component. 

The  general  equation  for  the  induced  e.m.f.  of  such  a  machine 
can  therefore  not  be  expressed  as  simply  as: 

E<  =  e  +  IZ  =  e  +  (i  +  ji,)(r  +  jx) 

=  e  +  ir  -  i&  +  j(ix  +  iir), 
but  must  be  written: 

Et  =  e  -f-  IT  —  i&i  -f  j(ix  +  z 
258 


CHARACTERISTICS  OF  ALTERNATORS          259 

where  x\  belongs  to  the  wattless  component  and  is  about  0.6z, 
and  x  is  the  reactance  belonging  to  the  power  component  of  the 
current. 

In  the  synchronous  impedance  test,  xif  is  obviously  determined 
since  the  current  in  that  test  lags  nearly  90  degrees  in  time  and 
hence  in  space.1  Similarly,  the  expression  for  the  field  excitation 
is  not 

Ff  =  —  mi  —  Cb  +  j(Ca  —  mil), 
but  is 

Ff  =  _  mi  -  Cb  +  j(Ca  - 


where  m  is  always  smaller  than  mi  since  m  determines  the  shift 
due  to  the  power  component  of  the  current.  The  relative 
values  of  m  and  mi  are  not  by  any  means  fixed  but  vary 
over  a  considerable  range.  It  may,  however,  be  assumed  that 
m  =  O.Swi. 

The  value  of  mi  is  determined  from  the  winding  data.  For 
instance,  in  a  three-phase  machine  it  is,  \/2  X  1.5  X  turns  per 
pole  and  phase. 

The  constants  for  a  definite  pole  machine  of  the  same  general 
dimensions  as  the  generator  previously  calculated  would  thus  be 

C  =  1,  mi  =  0.5,  r  =  0.02,  x  =  0.10, 
m  =  0.8  X  0.5  =  0.4,  Xl  =  0.6  X  0.10  =  0.06. 

The  angular  space  displacement  of  the  armature  with  reference 
to  the  field  structure  between  no-load  and  any  particular  load  is 
of  interest.  Consider  first  a  machine  with  round  rotor. 


FIG.  207. 

Let  e  be  the  Terminal  Voltage.  At  no-load  the  axes  of  the 
field  poles  are  in  the  directions  of  the  field  flux,  that  is,  in  direction 
oF0,  in  Fig.  207.  With  any  load,  01,  as  shown  in  the  figure,  the 
direction  and  the  magnitude  of  the  field  excitation,  the  former 

1  To  calculate  Xi  from  synchronous  impedance  test,  assume  i  =  0.  Then, 
substituting  in  (115),  Ff  =  -C6  +  j(Ca  -  wiii),  where  a  =  -to,  6  =  if. 


260 


ELECTRICAL  ENGINEERING 


assumed  the  same  as  the  direction  of  the  field  poles,  is  F/.  Thus 
the  angular  space  displacement  of  the  field  structure  in  reference 
to  the  armature  is  represented  by  the  angle  a,  and  tan  a  has 
been  shown  to  be 

mi  +  Cb 

tan  a  =  79  --  -' 
Ca  —  mi\ 

With  "  definite  pole"  machines  this  becomes, 

mi  +  Cb 


tan  a  = 


79 

Ca  — 


where,  of  course,  a  and  b  are  different  from  a  and  b  in  the  "round 
rotor"  case. 

As  an  illustration  consider  the  same  two  generators,  whose 
constants  have  been  given.  To  find  a  with  varying  power  factor. 

(1)  For  the  round  rotor  type  we  have: 

C  =  1;  m  =  0.5;  x  =  0.1;  r  =  0.02;  6  =  1; 

/  =  t  +  ji'  =  1 

a  =  e  +  ir  -  i'x  =  1  +  0.02*  -  0.1*' 
6  =  ix  +  *'r  =  0.1*  +  0.02^'. 

0.1*  +  0.02?:/  +  0.5*  0.6*  +  0.02*'  ^ 

=  1  +  0.02;  -  0.1*'  -  0.5*'  " 

Tabulating  : 


1  +  0.02*  -  0.6*' 


Leading  current 

Lagging  current 

P.F  

0.0 
0.0 
1.0 
0.0 

0.02 
0.02 
0.0 
-0.6 

0.4 
0.05 
2°  52' 

0.25 
0.25 
0.968 
0.15 

0.01936 
0.169 
0.005 
-0.5808 

0.425 
0.397 
21°  40' 

0.5 
0.5 
0.866 
0.3 

0.0173 
0.317 
0.01 
-0.519 

0.491 
0.645 
32°  50' 

0.75 
0.75 
0.662 
0.45- 

0.01324 
0.463 
0.015 
-0.3972 

0.618 
0.75 
36°  53' 

1.0 
1.0 
0.0 
0.6 

0.0 
0.6 
0.02 
0.0 

1.02 
0.588 
30°  27' 

0.75 
0.75 
-0.662 
0.45 

-0.013 
0.437 
0.015 
0.397 

1.412 
0.31 
17°  15' 

0.5 
0.5 
-0.866 
0.3 

-0.017 
0.283 
0.01 
0.519 

1.529 
0.185 
10°  30' 

0.25 
0.25 
-0.968 
0.15 

-0.019 
0.131 
0.005 
0.58 

1.585 
0.0826 
4°43' 

0.0 
0.0 
-1.0 
0.0 

-0.02 
-0.02 
0.0 
0.6 

1.6 
-0.0125 
-45' 

»' 

0.6i  

0  02i' 

«  
0  02i      ... 

-  0.6i'  

(  

tan  a  

(2)  For  machines  with  definite  poles. 
By  (112), 

Et  =  a  +  jb 
where  a  =  e  +  ir  -  i'x^  b  =  ix  +  i'r. 

By  (114),  < 

bC  +  mi 


tan  a  = 


—~ 

aC  — 


CHARACTERISTICS  OF  ALTERNATORS 


261 


=  0.4;  x  =  0.1;   xl 


In  this  case, 

C  =  1;   mi  =  0.5;   m  =  0.8 
0.6  a  =  0*)6 

r  =  0.02;  e  =  1;  /  =  i  +  #'  =  1. 
a  =  1  H-  0.02;  -  0.06t' 
6  =  Q.li  +  0.02z" 

O.U*  +  0.02^  +  QAi  0.5i  +  0.02z' 


tan  a  = 


s' 


1  +  0.02^  -  0.06^  -  0.56^       1  +  0.02z  -  0.56^  ~=  t1 


Leading  current 

Lagging  current 

P  F 

0.0 
0.0 
0.0 
1.0 

0.02 
0.02 
0.0 
-0.56 

0.44 
0.046 
2°  40' 

0.25 
0.25 
0.125 
0.968 

0.019 
0.144 
0.005 
-0.542 

0.463 
0.31 
17°  17' 

0.50 
0.50 
0.25 
0.865 

0.017 
0.267 
0.01 
-0.49 

0.527 
0.51 
26°  58' 

0.75 
0.75 
0.375 
0.662 

0.013 
0.388 
0.015 
-0.37 

.0.644 
0.60 
31°  6' 

1.0 
1.0 
0.5 
0.0 

0.0 
0.5 
0.02 
0.0 

1.02 
0.49 
26°  7' 

0.75 
0.75 
0.375 
-0.662 

-0.013 
0.362 
0.015 
0.37 

1.39 

0.26 
14°  35' 

0.50 
0.50 
0.25 
-0.865 

-0.017 
0.233 
0.01 
0.49 

1.50 
0.155 

8°  49' 

0.25 
0.25 
0.125 
-0.968 

-0.019 
0.106 
0.005 
0.542 

1.55 

0.068 
3°  53' 

0.0 
0.0 
0.0 
-1.0 

-0.02 
-0.02 
0.0 
0.56 

1.56 
-0.0127 
-45' 

0  5i 

V         

0.02^  .  . 
s'             

0.02i  
-  O.SGi'  

tr     

tan  a 

The  curves  for  both  machines  are  plotted  in  Fig.  208. 


0.25         0.5         0.75 
Leading 


0.75         0.5         0.25 

Lagging 
Power  Factor 


1.0 
wer 

FIG.  208. 

Let  it  be  required  to  solve  the  following  problem: 

Problem  90. — An  alternator  has  the  rating: 
A.T.B.  8-100-900-2300  volts  Y. 


262  ELECTRICAL  ENGINEERING 

The  saturation  curve  is  given  by  the  following  data : 


A.T. 

• 

A.T. 

e 

400 

250 

3650 

2300 

1000 

700 

5000 

2700 

2000 

1470 

6000 

2900 

3000 

2060 

The  normal  gap  density  is  40,000  and  the  average  gap  length  is  0.25  in. 
The  armature  reaction  is  1490  A.  T.  per  pole. 

The  synchronous  impedance  test  gives  1890  A.T.  excitation  with  full-load 
current  at  0.25-in.  average  gap,  and  1990  A.T.  with  0.1875-in.  average  gap. 

The  armature  resistance  per  phase  is  0.69  ohm. 

The  weight  of  the  revolving  element  is  800  lb.;  the  radius  of  gyration  is 
0.86ft.1 

First.  Draw  the  saturation  curves  with  0.25-in.  gap  (data  given  above) 
and  also  with  0.1875-in.  gap. 

Second.  Determine  m,  m^  x,  Xi  both  for  round  rotor  and  definite  pole 
machines. 

Third.  Draw  the  curve  of  field  excitation,  Ff,  for  varying  non-inductive 
load  /,  (compounding  curve),  for  the  two  round  rotor  and  the  two  definite 
pole  machines. 

Fourth.  Calculate  and  plot  the  terminal  voltage  as  the  non-inductive  load 
is  increased  and  the  field  excitation  kept  at  the  normal  no-load  excitation. 

Use  0.25-in.  gap  and  0.1875-in.  gap  with  the  two  types  of  machines. 

Solution.—  First.  From  the  saturation  curve  data,  the  total  A.T.  at 
2300  volts,  for  0.25-in.  gap  =  3650. 

Thus,  3650  =  gap  A.T.  +  iron  A.T. 

From  Eq.  (12),  gap  amp.-turns  =  0.313  X  flux  density  X  gap  length. 

/.for  0.25-in.  gap,  gap  A.T.  =  40,000  X  0.313  X  0.25  =  3130;  for  0.1875-in. 
gap,  gap  A.T.  =  40,000  X  0.313  X  0.1875  =  2347.  The  iron  A.T.  are  the 
same  for  both  gaps. 

To  plot:  (1)  plot  the  given  curve;  (2)  draw  the  straight  line  gap  satura- 
tion (magnetization  curve)  for  0.25  in.,  through  0  and  3130  at  2300  volts; 
(3)  draw  the  straight  line  saturation  for  0.1875  in.  through  0  and  2347  at 
2300  volts;  (4)  add  to  (3)  the  difference  between  (1)  and  (2). 

The  curves  are  given  in  Fig.  209. 

Second.     To  determine  m,  mi,  x,  XL 

(a)  Definite  pole  machine;  air  gap  =0.25  in. 

From  p.  259,  Chap.  XXXV, 


arm.  reaction       1490       1490 


arm.  current 


since 


/  = 


100,000 
V3  X  2300 


25 


=  25  amp. 


=  59.5, 


Data  for  later  calculation  of  hunting.     See  Chap.  XXXVIII. 


CHARACTERISTICS  OF  ALTERNATORS          263 


Assuming 


m 


m  =  0.8  X  59.5  =  47.6. 
To  determine  o?i.     Under  short-circuit  test  e  =  0  and,  in  (112)  therefore, 


where  /  is  the  effective  current. 
And 

Fr  =  jCIr  -  CIxi. 


3000 


(3)  (2) 


2200 


1800 


1400 


1000 


200 


7 


r 


17 


7 


z 


(4) 


0  1000  2000  3000  4000  5000 

Ampere-Turns 
FIG.  209. 

Since  the  current  lags  nearly  90°,  /  =  -  jii,  and,  neglecting  r  in  compari- 
son with  x\j  this  last  equation  may  be  written.1 

Fr  =  -  Clxt  =  -  Cixi  +  jdiXi  =  Ff  +  Fa 
where  ^o  =  —  fWiJti« 

/.  Ff  =  j(CiiXi  +  mil) 

and  Ff  =  diXi  -\-  miii 

whence 

(116) 


Cii 


1  When  r  is  not  neglected,  Xi  = 


264  ELECTRICAL  ENGINEERING 

Supplying  values: 

(a)  Definite  pole  machine,  0.25-in.  gap 

Ff   =  1890 
ilml  =  25  X  59.5  =  1490, 


"  1330 

_  1890  -  1490 
•  •  *i  =     2.74  X  25 

and 

5.84 

*  -  06    =  9'5' 

(6)  Round  rotor  machine,  0.25-in.  gap. 
m  =  59.5,  being  the  same  as  mi  in  definite  pole  machine. 
x    =    9.5,  being  the  same  as  x  in  definite  pole  machine. 

(c)  Definite  pole  machine,  0.1875-in.  gap. 

wi  and  m  are  the  same  as  for  the  0.25-in.  gap,  since  the  armature  ampere 
turns  are  independent  of  the  gap. 

.*.  mi  =  59.5,  m  =  47.6. 
1990  -  1490 


since 

9870 
Ff  -  1990,  and  C  =    ~  =  2.16. 

9.25 


(d)  Round  rotor  machine,  0.1875-in.  gap. 

m  =  59.5 
x    =  15.4. 

Third.     Compounding  curves,  —  (/'/vs.  7).     (a)  Round  rotor   machine; 
gap  =  0.25  in.     Non-inductive  load.     By  (115), 


Ff  =  V(-  bC  -  mi)*  +  (aC  -  mil)2, 
where 

3650 
=  1330  =  2-'742»1  m  =  59-5J  i  =  variable; 

ii  =  0;  x  =  9.55;  r  =0.69;  e  =  1330 
a  =  e  +  ir  -  i&  =  1330  +  0.69i 

b  =  ix  +  iir  =  9.55i. 
Whence, 

Ff  =  V(-  9.55i  X  2.742  -  59.5i)2  +  ((1330  +  0.69i)  2.742)2 


=  \13.3  X  106  +  13,800i  +  7354i2 

=  85.75  Vi2  +  1.875*  +  1802  =  85.75  vT. 

1  C  is  constant  only  on  the  assumption  of  a  straight  line  magnetization 
curve. 


CHARACTERISTICS  OF  ALTERNATORS 

Tabulating : 


265 


I 

0 

10 

20 

25 

30 

40 

50 

1.875*  

o 

18  75 

37   ^ 

4fi  S^ 

K(\     0 

•5  c  fi 

i2 

o 

100 

400 

fioe 

onn 

/o.  u 
i  Ann 

93.6 

t  

1802 

1921 

2239 

2474 

97  KQ 

1OUU 

1477 

2500 

1  '•*(!»' 

Vf  

Ff  

42.5 
3650 

43.8 
3760 

47.25 
4050 

49.6 
42^0 

52.5 

4.crjc 

58.9 

Cftcrv 

loyo 
66.2 

CA*rr 

oo/o 

(6)  Round  rotor;  gap  =  0.1875  in.     Non-inductive  load. 

2870 
C  =  j7j3Q  =2.16;  x  =  15.38;  6  =  15.38*. 

Other  quantities  are  as  in  (a). 

•'•  Ff  =  V(  -  15.38*  X  2.16  -  59.5*)2  +  ((1330  -f  0.69*)  2.16)2 


92.7  Vi2  +  0.995*  +  960  =  92.7  \/t . 


Tabulating : 


i 

0 

10 

20 

25 

30 

40 

50 

0.995*  

0 

9.95 

19  9 

24  95 

29  85 

39  8 

49  75 

t  

960 

1070 

1380 

1610 

1890 

2600 

3510 

V7  

31 

32.7 

37.1 

40.1 

43  5 

51  0 

59  2 

Fj  

2870 

3030 

3440 

3720 

4030 

4730 

5490 

(c)  Definite  pole  machine;  gap  =  0.25  in.     Non-inductive  load. 


bC  -  mi)2  +  (aC  - 

where  C  =  2.742;  mi  =  59.5,  m  =  47.6;  i  =  variable;  ii  =  0;  e  =  1330; 
x  =  9.55;  b  =  ix  =  9.55i;  a  =  e  +  ir  —  i&i  =  1330  +  0.69i. 

/.  Ff  =  V(-  9.55i  X  2.742  -  47.6i)2+  ((1330  +  0.69i)  2.742)2 


=  73.8  \/*2  +  2.53i  +  2430  =  73. 
Tabulating  : 


i 

0 

10 

20 

25 

30 

40 

50 

2.53i  

0 

25.3 

50.6 

63.25 

75.9 

101.2 

126.5 

t  

Vt 

2430 
49  3 

2555 
50.5 

2881 
53.6 

3118 
55.8 

3406 
58.3 

4131 
64.3 

5057 
71.0 

Ff. 

3650 

3735 

3960 

4120 

4310 

4750 

5250 

(d)  Definite  pole  machine;  gap  =  0.1875  in.     Non-inductive  load. 

C  =  2.16;  X  =  15.38;  m  =  47.6;  6  =  15.38*. 
/.  Ff  =  V(-  15.38*  X  2.16  -  47.6i)2  +  ((1330  +  0.69*)  2.16)2 
=  80.8  Vi2  +  1.3H  +  1265  =  80.8  \/T. 


266  ELECTRICAL  ENGINEERING 

Tabulating: 


i 

0 

10 

20 

25 

30 

40 

50 

1.3K  

0 

13.1 

26.2 

32.75 

39.3 

52.4 

65.5 

t  

1265 

1378 

1691 

1923 

2204 

2917 

3831 

VT  

35.5 

37.1 

41.1 

43.8 

47.0 

54.0 

61.9 

Ff  

2870 

2995 

3320 

3540 

3800 

4360 

5000 

10  20  30  40  50 

Armature  Current 

FIG.  210. 


Compounding  curves  for  all  four  machines  are  given  in  Fig.  210. 
Fourth,     (a)  Round  rotor  machine  ;  gap  =0.25  in. 
At  no-load,  the  induced  voltage  to  neutral  is  e0  =  1330. 
The  field  ampere-turns  are  E/  =  3650. 
Eq.  (115)  may  be  written 

Ft  =  V[-  C(i  x  -Hif)  -  mi]z  +  [C(e  +  ir  -  iix)  -  mil]2 
or,  since  ii  =  0 


Ff  =  V[  -  Cix  - 


[C( 


CHARACTERISTICS  OF  ALTERNATORS          267 


Numerical  values  previously  found  are: 

C  =  2.742;  x  =  9.55;  m  =  59.5;  r 

.".  Ff  = 


0.69. 


-2.742  X9.55z  -  59.5z]2  +  [2.742(e  +  0.69i)]2 
=  \/7.54e2  +  10.4ei  +  7364i2  =  3650. 
Squaring  both  sides  and  reducing,  gives 

(Ff)*  =  e2  +  1.38ei  +  977i2  =  1.763  X  10« 
whence 

Tabulating : 


e  =  -  0.69i  ±  31.25  V1805  - 


i 

0 

10 

20 

25 

30 

40 

50 

-0.691...  
»i 

0 
0 

-  6.9 
100 

-13.8 
400 

-17.25 
625 

-20.7 
900 

-27.6 
1600 

-34.5 
2500 

1805  -  i2 

1805 

1705 

1405 

1180 

905 

205 

-695 

V1805  -  i2  

31.25V  
e 

42.5 
1330 
1330 

41.25 
1290 
1283 

37.45 
1170 
1156 

34.35 
1073 
1056 

30.05 
940 
919 

14.3 
447 
420 

\/3e  

2300 

2220 

2000 

1830 

1590 

726 

(6)  Round  rotor  machine;  gap  =  0.1875  in. 
The  field  ampere-turns  are  Ff  =  2870. 

Ff  =  V[-  Cix  -mi]2  +  [C(e  +  ir)]2, 


where 


C  =  2.16;  x  =  15.38;  m  =  59.5;  r  =  0.69. 

Ff  = 


-  2.16  X  15.38i  -  59.5i]2  +  [2.16(fl  +  0.69i)]2 
=  2870 


whence 

e 
Tabulating : 


=  \4.68e2  +  6.44ei 

=  -  0.69*  ±  42.8  \/960  -  i2, 


i 

0 

10 

20 

25 

30 

40 

50 

-0  69i 

0 

-6.9 

-13.8 

-17.25 

-20.7 

-27.6 

-34.5 

i2  

0 

100 

400 

625 

900 

1600 

2500 

960  -  i2  

$60 

860 

560 

335 

60 

-640 

-1540 

V960  -i2  

42.8\/~  
e 

31 
1330 
1330 

29.3 
1260 
1253 

23.65 
1013 
999 

18.3 
783 
766 

7.74 
331.5 
311 

\/3e  

2300 

2170 

1730 

1325 

538 

(c)  Definite  pole  machine;  gap  =  0.25  in. 

Ff  =  V[-  Cix  -  mi]2  +  (C(e  +  ir)]2 
where 

C  =  2.742;  x  =  9.55;  m  =  47.6;  r  =  0.69. 


_ 

Ff  =  \/[-  2.742  X  9.55J  -  47.6J]2  +  [2.742(e  +  0.6&)]2 
=  V7.54e2  +  10.4  ei  +  546417"2  =  3650 


268 
whence 


ELECTRICAL  ENGINEERING 


e  =  -0.69i  ±  26.9  V  2430  -  i2. 
Tabulating: 


•i 

0 

10 

20 

25 

30 

40 

50 

-0.69i  

0 

-6.9 

-13.8 

-17.25 

-20.7 

-27.6 

-34.5 

i*      

0 

100 

400 

625 

900 

1600 

2500 

2430  -i2  
V2430  -  i2  
26.9\/~~ 

2430 
49.25 
1330 

2330 
48.25 
1300 

2030 
45 
1212 

1805 
42.5 
1144 

1530 
39.1 
1053 

830 
28.8 
775 

-70 

e         .... 

1330 

1293 

1198 

1127 

1032 

747 

\/3e 

2300 

2240 

2072 

1950 

1785 

1293 

(d)  Definite  pole  machine;  gap  =  0.1875  in. 

C  =  2.16;  x  =  15.38;  m  =  47.6;  r  =  0.69 

'[-  2.16  X  15.38i  -  47.6i]2  +  [2.16(e  +  0.69i)]2 


whence 


=  \/4.68e2  +  6.44ei  +  6532T2  =  2870 
=  -  0.69*  +  37.3  Vl265  -  iz. 


2400 


10 


20  80  40 

Load  Current,  Amperes 

FIG.  211. 


50 


CHARACTERISTICS  OF  ALTERNATORS 

Tabulating : 


269 


i 

0 

10 

20 

25 

30 

40 

50 

-069i 

0 

-6.9 

-13.8 

—  17.25 

-20  7 

-27  6 

-34  5 

i2 

o 

100 

400 

625 

900 

1600 

2500 

1265  -  i2  

1265 

1165 

865 

640 

365 

-335 

-1235 

A/1265  -i2  
37.3\/~~  
e  
A/3e  

35.5 
1330 
1330 
2300 

34.1 
1272 
1265 
2190 

29.4 
1097 
1083 
1875 

25.3 
944 
927 
1605 

19.1 
712 
691 
1196 

The  curves  of  terminal  voltage  with  varying  non-inductive  load  are  shown 
plotted  in  Fig.  211. 

Problem  91. — Determine  all  the  above  quantities  and  plot  the  curves  for 
the  conditions  of  load  power  factor  of  80  per  cent.,  both  lagging  and  leading. 

Problem  92. — Carry  out  the  above  investigation  for  the  same  alternator 
when  the  phases  are  delta-connected. 


CHAPTER  XXXVI 


Pole 


APPROXIMATE  DETERMINATION  OF  THE  SELF-INDUC- 
TION OR  LOCAL  MAGNETIC  LEAKAGE  REACTANCE 
OF  AN  ALTERNATOR 

Consider  a  single  slot  of  an  armature  situated  directly  under- 
neath a  pole.  When  current  flows  in  the  winding,  lines  of  flux 
are  set  up,  linking  with  the  turns  of  wire.  Thus,  as  in  Fig.  212, 
some  lines  will  pass  across  the  space  of  width,  a,  occupied  by  the 
conductors;  some  will  pass  across  the  width,  c,  of  the  upper  insu- 
lation; some  across  width,  d,  which  is 
occupied  by  the  wedge;  and  some  will 
pass  across  the  face  of  the  tooth,  /, 
and  air  gap,  g,  into  the  pole.  Each 
of  these  fluxes  is  set  up  by  a  mag- 
netomotive force  acting  through  a 
reluctance,  the  values  of  both  of 
which  may  be  determined  in  each 
case.  The  reluctance,  and  conse- 
quently the  flux  and  the  inductance 
are,  in  the  following,  first  determined 
per  centimeter  length  of  effective  iron 
parallel  to  the  shaft.  The  inductance  of  the  end  connections  or 
the  parts  of  the  coil  which  are  outside  of  the  iron  is  considered 
separately. 

1.  Inductance,  LI,   Due   to  Flux    through    Section  a. — The 

x 

magnetomotive  force  is  FI  =  nl  -?  where  n  is  the  total  number  of 

a 

turns  in  the  slot  and  I  is  the  current;  -  is  any  portion  of  the 

depth  of  the  coil,     n  -  gives  the  number  of  turns  included  in  any 
distance  x  from  the  bottom  of  the  coil.     The  flux  set  up  by  this 

47rn7- 

a  b 

m.m.f.,  FI,   is  d<f>i  =  -     —i  where  pi  =  -r-  =  reluctance  of  a 
Pi  dx 

270 


DETERMINATION  OF  THE  SELF-INDUCTION   271 

small  path  of  length  6,  in  air,  and  of  cross-section,  dx  sq.  cm. 
(dx  X  1).     The  reluctance  of  the  iron  is  neglected.     Then, 

x  dx 
-  -=-• 
a  o 

The  interlinkages  or  turns  linking  with  this  flux  are  n-  •     Hence, 

the  flux-turns  interlinkage  per  unit  current,  or  the  inductance 
across  the  width,  a,  of  the  coil,  is 

1  Ca     n2    x2  a      n2 

Li  =  j  I    4ir-^I  -^dx  =  g47ry,  per  cm.  length  of  effective  iron. 

2.  Inductance  Due  to  the  Flux  through  Section  c.  —  The  mag- 
netomotive force  is  F2  =  nl.     p2  =  — 

The  flux  is 

c 


P2  0 

All  the  n  conductors  link  with  this  flux. 


3.  Inductance  Due  to  the  Flux  across  the  Section  d.  —  This,  by 
a  similar  process,  is 


e 


4.  Inductance  over  the  Face  of  the  Tooth.  —  The  magneto- 
motive force  is  F4  =  nl.  The  reluctance,  p\  =  -r-  The  flux 
set  up  is 


All  the  n  conductors  link  with  this  flux. 


5.  Inductance  of  End-connections.  —  The  inductance  of  the 
part  of  the  winding  that  projects  outside  of  the  iron  is  almost 
impossible  to  estimate  accurately.  It  depends  largely  on  the 
mechanical  design.  If  the  end  shields  are  some  distance  away 
from  the  winding,  a  fair  approximation  is  obtained  by  assuming 
the  flux  per  ampere-turn  per  centimeter  of  wire  to  be  inversely 
proportional  to  the  square  root  of  the  pole  pitch,  or  what  is 
equivalent,  the  square  root  of  the  armature  diameter  divided  by 


272  ELECTRICAL  ENGINEERING 

the  number  of  poles.     A  fair  approximation  to  the  flux  per  ab- 
solute ampere-conductor,  per  centimeter  of  wire  is 


'5  =  13\JY)  ( 


Y)  (or  1.3  A       when  amperes  are  used), 

where  p  is  the  number  of  poles,  and  D  =  armature  diameter. 
If  the  length  of  the  end-connections  of  a  coil,  counting  both 

sides  of  the  core,  is  8  X  —  >  then  the  flux  per  turn,  per  ampere- 
conductor, is 


Thus,  with  a  single-coil  winding,  where  all  conductors  of  a 
coil  are  wound  together,  the  flux  per  coil  is 


05  =  104n/  X  \—  > 
Mp 

where  n  is  the  number  of  effective  conductors  and  /  is  the  current 
per  conductor  in  absolute  amperes. 

The  inductance,  L5,  will  be  due  to  only  one-half  of  this  flux 
since  each  coil  occupies  two  slots. 


.-.  Lb  =  ^-  =  52n< 


ID 
A/— • 

\  p 


The  total  inductance  for  a  single  slot  exclusive  of  end-connec- 
tions is  then,  per  centimeter  net  length  of  armature  iron, 


and  the  total  inductance,  including  end-connections  for  length, 
I,  of  iron,  is,  in  henrys, 

a    ,   c   ,   d   ,    /        13     ID 


109  ,__. 

For  a  three-phase  alternator, 
with  one  slot  per  pole  per  phase, 
there  is  also  to  be  added  a  term 
due  to  the  flux,  06,  in  parallel 
with  04,  which  passes  from  the 
next  adjacent  half  tooth,  across 
the  gap  (Fig.  213). 
FIG.  213.  The  inductance  due  to  this  flux 

will  vary  greatly,  according  to  the 

air  gap,  whose  cross-section  and  length  may  be  very  different 
from  the  values  used  in  determining  L4. 


DETERMINATION  OF  THE  SELF-INDUCTION   273 

Problem  93. — Calculate  the  reactance  per  phase  of  the  following  alter- 
nator when  the  slots  are  under  the  poles. 

A.T.B.  8-100-900-2300  v. 

48  slots;  24-in.  armature  diameter;  therefore,  2  slots  per  pole  and  phase ;  28 
effective  conductors  per  slot.     Dimensions,  referring  to  Fig.  214,  are: 

a  =  1 . 0  g    =  average  gap  under  adjacent  teeth  =  0.25 

6  =  0.75  g'  =  average  gap  under  distant  teeth  =  0.5 

d  =  0.14  n   =  28 

e  =  0.27  s     =  2 

/  =  0.10  p   =  8 

h  =  0.85  I    =  9 

k  =  0.82  D  =  24 


The  wires  are  confined  to  the  distance,  a,  of  the  slot. 

Each  slot  has  n  effective  conductors  and  there  are  s  slots  per 
pole  per  phase.  Let  0i  be  the  field  which  crosses  the  conductors 
due  to  the  m.m.f.  of  the  coil  which  is  between  the  bottom  of  the 
slot  and  the  distance  X. 

Then  the  m.m.f.  is 

*—,£/, 

where  7  is  the  current  in  amperes  in  the  conductor. 

The  flux,  in  section  dx  per  cm.  depth  of  magnetic  circuit 
parallel  to  the  shaft,  is  therefore, 

4?rFi  _  birsnxl 
Pi  Pi® 

But  the  reluctance  pi  of  the  path  is 

Pi  =  -r-»  neglecting  the  iron. 

18 


274  ELECTRICAL  ENGINEERING 

Thus,  4wsnxldx       4:wnlxdx 

dd>i  =  r =  T * 

asb  ab 

M 

This   flux   interlinks  with  -  cs  conductors.     Therefore   the 

inductance  of  this  part  of  the  magnetic  circuit,  which  is  the 
interlinkages  of  the  turns  and  flux  across  the  conductors  per 

unit  current,  is 

^a 

4.Trn2s  T  „  7  a 

Ix2dx  =  47rn2s  ^r- 
3o 

Consider  next  the  inductance  of  the  part  of  the  magnetic 
circuit  which  is  above  the  coil  proper. 

The  m.m.f.  is  that  of  all  conductors,  and  is  Fz  =  snl. 
The  flux  02,  per  cm.  length,  is 


Pz  P2 

In  this  particular  case  there  are  three  magnetic  paths  in  multi- 

ple, the  first,  of  reluctance,  -j>  the  second,  of  reluctance,  —  >  and 

(t  6 

sb 
the  third,  of  reluctance,  y 

•    1  ;;   1    .    1         1    _d+f       e_ 
*  "    2       sb       sh       sb          sb          sh 


and 

d+f 


Some  flux  crosses  the  two  gaps  from  the  teeth  adjacent  to  the 
coils  and  causes  an  inductance  which  is  similarly  determined. 
Thus,  the  m.m.f.  is 

F3  =  snl. 

4irsnl  k 


PS  2g 

•7  1A  T    k 

.  .  L3  =  j  4TrsnI  pr-  sn 
L  zg 


DETERMINATION  OF  THE  SELF-INDUCTION   275 

Similarly,  the  flux  which  crosses  the  gaps  from  the  more  dis- 
tant teeth  causes  an  inductance, 

L4  =  47Tsn2    -r 


Thus,  the  total  inductance  per  centimeter  depth  of  magnetic 
circuit  covered  by  the  iron  is 

T  t  9  F  a    ,   d  -f  /  ,   e    ,    ks   ,    ks~\ 

L°  =  4?rm2  L§5  +  V  +  1  +  *,  +  w\' 

If  I  is  the  net  length  in  centimeters  of  the  iron  of  the  armature 
core,  and  p  is  the  number  of  poles,  then  the  inductance  per  phase 
of  the  part  of  the  electric  circuit  which  is  in  the  slots  is 

,  7r  a       d  +  /  .    e    .    ks   .    ks~\ 

Lo  =  4r«.«pl  [36  +  V  +  h  +  2g  +  W\' 

(The  dimension  of  inductance  and  capacity  in  the  absolute  system 
of  units  is  centimeters.) 

By  extending  the  reasoning  in  the  case  of  a  single  slot,  the  in- 
ductance of  the  end-connections  per  phase  is  found  to  be 


L5  =  52sWp\  —  =  52s2ra2 
\  p 

With  bar  winding,  when  the  coils  are  split  up,  as  shown  in 
Fig.  215,  the  inductance  of  the  end-connections  becomes  L5  = 

77  9  / 

13s2n2  \/Dp,  since  the  m.m.f.  per  end  coil  is  —^-t  and  the  inter- 
linkages  are  ^-     The  inductance  of  the  ma-     /    \ 

chine  per  phase  is  then 

L  =  Lo  +  L6  in  cm.,  or 

•r         LO  ~r  LS  , 

L  =  — ^ —  henrys. 

If  inch  measurements  are  used,  FIG.  215. 

f  .   e    .   ks 


and 

L5  =  83s2n2  \/Dp  for  single  coil  winding,  or 

=  20.8s2n2  \/T)p  for  split  coils,  and 
the  inductance  in  henrys  is 

L  = 


276 


ELECTRICAL  ENGINEERING 


Applying  these  equations  to  the  particular  three-phase  alter- 
nators given  above,  and  noting  that  the  coils  are  not  split  up,  we 
get: 

Lo  =  32  X  2  X  282  X  8  X  0.9  [3^75  +  ^  +  5^  + 

(0.445  +  0.32  +  0.33  + 

o^1_2  +  q:^x_2]  =  21)600j000cinj 

3.27  +  1.64) 


and 


and 


L5  =  83  X  4  X  282  X  Vl92  =  3,600,000  cm. 
.'.  L  =  25>20^000  henrys  =  0.0252  henrys. 


x  =  27r60  X  0.0252  =  9.5  ohms. 
This  is,  then,  the  reactance  for  the  slot  under  the  pole,  that  is, 

the  reactance  which  should  be  used  with  the  power  component 

of  the  current.  The  reactance  be- 
tween the  poles  is  less  and  may  be 
taken  as  0.6z  or  xi  =  5.8  ohms. 

It  is  very  convenient  in  designing  a 
slot,  to  make  it  accommodate  four 
coils.  As  this  is  a  very  common 
arrangement,  the  calculation  of  the 
inductance  of  a  single  tooth  armature 
having  four  coils  in  the  slot  is  also 
made.  The  cross-section  of  the  slot 
is  shown  with  dimensions  in  Fig.  216. 

The  procedure  is  practically  the  same  as  in  the  preceding  case. 

The  flux  through  a  small  section,  dx,  of  the  space  occupied  by 

the  lower  pair  of  coils  is 


x  N 
where  the  magnetomotive  force  is  Fx  =  —  -^I,N  being  the  total 

(i    & 

number  of  turns. 

Thus,  the  flux-turns  interlinkage  per  unit  current  or  the  induct- 
ance through  the  lower  pair  of  coils,  LI,  is 


dx 


ba 


DETERMINATION  OF  THE  SELF-INDUCTION   277 
The  inductance  across  the  insulation,  h,  between  the  layers  is 

ik?» 

L2  =  ^_4         _          h 
b  b 

The  inductance  across  the  upper  coils  is 


1  f°4'TJ1+g 

/JL     -r 


a 


b 


a  _  SrrJV^a  ^  TrN2a  _  7irN2a 
o  ~      36  36  36 


The  inductance  across  the  insulation,  c,  beneath  the  wedge,  is 


The  inductance  across  the  wedge  is  L5  = 
The  inductance  across  the  face  of  the  tooth  is 


The  inductance  of  end-connections  is,  as  in  the  previous  case, 

L7  =  527V2  J—  • 

\  p 

The  total  inductance  per  centimeter  effective  length  of  core  is 


Problem  94.  —  A  certain  three-phase,  60-cycle  alternator  has  one  slot  per 
pole  per  phase.  The  dimensions  in  inches  of  slot,  etc.,  are  as'  in  Fig.  216, 
where  a  =  0.45,  6  =  0.75,  c  =  0.14,  d  =  0.37,  e  =  0.85,  /  =  0.82,  g  =  0.15, 
h  =  0.1.  There  are  twenty-four  slots,  thirty-two  effective  conductors,  the 
effective  length  of  the  armature  core  is  9  in.,  the  armature  diameter  is  32 
in.  Show  that  the  armature  reactance  is  approximately  4  ohms. 


CHAPTER  XXXVII 


FIG.  217. 


ARMATURE  REACTION  IN  MULTIPHASE  MACHINES 

With  current  in  the  armature  of  an  alternator,  two  magneto- 
motive forces  exist,  one,  that  of  the  field  winding,  and  the  other, 

that  of  the  armature  winding. 

Sometimes  these  add  directly 
but  more  often  they  are  more  or 
less  in  opposition. 

If  the  resultant  field  flux  is  in 
the  direction  of  the  field  poles, 
Fig.  217,  and  the  armature  winding 
is  assumed  concentrated  in  a  coil 
in  position  a-b,  then  the  induced  e.m.f.  due  to  the  rotation  of  the 
coil  in  the  field  is 

e{  —  Em  sin  0 

and  the  current  is 

i  =  Im  sin  (6  +  a), 

where  a  is  the  angle  of  lead  of  the  current  in  respect  to  the  e.m.f., 
that  is,  tan  a  =  -,  where  x  and  r  are  the  total  reactance  and  re- 
sistance of  the  external  and  armature  circuits,  and  EM  and  IM  the 
maximum  values  of  the  e.m.f.  and  current  respectively. 

Jf  the  armature  coil  has  T  turns,  the  m.m.f.  of  the  armature  is 
obviously, 

iT  =  JJTsin  (B  +  a), 

In  the  position  shown  the  m.m.f.  of  the  armature  does  not  act 
in  line  with  the  m.m.f.  of  the  field  winding,  but  its  component  in 
the  direction  of  the  field  is  a'  -  &'  or  the  total  m.m.f.  multiplied 
by  cos  0. 

The  component  b  -  b'  of  the  armature  m.m.f.  at  right  angles 
to  the  field  is,  of  course,  the  total  m.m.f.  multiplied  by  sin  B 
But  this  component  does  not  increase  or  decrease  the  field,  but 
only  distorts  it. 

278 


ARMATURE  REACTION  279 

Let  M  be  the  component  of  the  armature  m.m.f.  in  the  direc- 
tion of  the  field  m.m.f.     Then 

M  =  ImT  sin  (0  +  a)  cos  0 

=  ImT  (sin  B  cos  a  +  cos  0  sin  a)  cos  0 
=  ImT(%  sin  20  cos  a  +  cos2  0  sin  a). 
But 

COS2  e  =  L±-™*°. 

/mT7 

•'•  M  =  ~T~  tsin  2e  cos  a  +  sin  a  cos  20  +  sin  «1 

7  T 

=  -y-  [sin  (20  +  a)  +  sin  a]. 

It  is  seen  that  the  average  value  of  the  armature  reaction  in  the 

T     rp 

direction  of  the  poles  has  a  constant  value  which  is  -^-  sin  a, 

2i 

and  superimposed  upon  this  is  a  pulsating  reaction,  a  m.m.f. 
which  pulsates  at  double  frequency.  The  effect  of  the  latter  is 
zero  when  considering  the  average  effect  over  a  cycle. 

T    rp 
IT  J-mJ- 

.  .  Mav.  =  — y  sm  a, 

But  Im  sin  a  is  the  maximum  value  of  the  wattless  component 
of  the  current  (Fig.  218). 

Thus  the  armature  m.m.f.  (or  armature  reac- 
tion,  as  it  is  called),  in  the  direction  of  the  poles 
corresponds  to  the  wattless  component  of  the          FIG.  218. 
current. 

Thus,  if  the  current  is  in  time-phase  with  the  induced  e.m.f. 
(in  which  case  there  is  no  wattless  component),  the  armature 
current  neither  magnetizes  nor  demagnetizes  the  field,  but  only 
distorts  the  distribution  of  the  flux. 

If  the  armature  current  leads  the  induced  e.m.f.,  then  it  is  seen 
that  the  armature  reaction  is  positive.  It  helps  the  field  m.m.f. 

If  the  current  lags,  then  a  is  negative  and  the  armature  reac- 
tion opposes  the  field  m.m.f. 

In  a  three-phase  machine  the  e.m.fs.  of  the  different  phases 
may  be  expressed  as 

ei  —  Em  sin  0 

e2  =  Em  sin  (0  +  120) 

e3  =  Em  sin  (0  +  240). 


280  ELECTRICAL  ENGINEERING 

Prove  that  the  average  armature  reaction  in  the  direction  of 
the  poles  is  1.57  mT  sin  a,  and  is  not  pulsating  but  steady. 

NOTE.  —  In  specifications  of  alternators  one  item  is  usually  called  armature 
reaction  and  the  value  given  is  ~  o~>  in  a  single-phase  machine,  ImT  in  a 

two-phase  machine,  and  l.5ImT  in  a  three-phase  machine. 

In  this  case,  however,  Im  is  the  maximum  value  of  the  rated  current,  and 
T  is  the  effective  number  of  turns  per  armature  pole  per  phase. 

Example.  —  Find  the  so-called  armature  reaction  in  an  8-pole, 
100-kw.,  2300-  volt,  three-phase  generator  which  has  224  armature 
turns  per  phase  and  which  is  Y-connected. 

Answer.  —  The  voltage  per  phase  is 

=  1330. 


The  full-load  effective  current  is 

100,000 

—^=-  -  =  25.1  amp. 

\/3  X  2300 

.'.  Im  =  25.1  \/2  =  35.5  amp. 

The  winding  is  practically  concentrated  so  that  all  turns  are 
effective,  thus 


.'.  Ma  =  1.5  X  35.5  X  28  =  1490  A.T., 

and  this  is  the  numerical  value  given  to  "  armature  reaction." 

If  the  armature  actually  carried  full-load  current  and  the  cur- 
rent was  lagging  90  time  degrees  behind  the  e.m.f.,  and  hence  was 
90  space  degrees  displaced  from  the  main  field  flux  then  the  de- 
magnetizing ampere-turns  would  be  1490. 

If  the  current  was  leading  then  the  armature  current  would 
assist  the  field  to  the  extent  of  1490  A.T. 

With  a  phase  angle,  say  30°,  the  actual  magnetizing  or  demag- 
netizing ampere-turns  would  obviously  be  only  745. 

In  an  n-phase  machine  the  armature  reaction  is  not  pulsating 
but  has  a  constant  value, 

M  Im™      • 

Ma    =    -  ^  -    Sm    a' 

Consider  any  particular  phase  indexed  m. 
Its  voltage  is 

e  =  Em  sin   (0  +  -—  J  ; 


ARMATURE  REACTION 


281 


its  current  is 

;  =  7m 
its  m.m.f.  is 

M  =  iT  =  7raT  sin 


cos 


The  total  m.m.f.  at  any  instant  is,  thus 


But, 


m  =  n 

M  =     S 

m  =  1 


in  (20  + 


sin 


+ 


=  sin  (26  +  a)  cos 


cos  (20  +  a)  sin 


The  sum  of  all  terms  containing  cos must 


n 


be  zero,  because 
sides  in  a  closed 


the  sum  of  the  cosines  of  all 
polygon  is  zero.     Similarly  the  .. 

are  zero.     Thus  it  fol- 


sn 


n 


terms  containing 

lows  that, 

Ma  =  M  (since  it  is  constant  for  all  values  of  0)  =  n 


FIG.  219. 


sin  a. 


FIG.  220. 


Effect  of  Distributed  Winding  on  the 
Armature  Reaction. — Consider  a  single- 
phase  armature  wound  with  a  number  of 
coils  as  is  shown  in  Fig.  220,  6,  all  of  whose 
coils  are  connected  in  series. 

The  effective  value  of  the  e.m.f.  gener- 
ated in  coil  A  may  be  represented  by.OA. 
The  e.m.f.  in  coil  B  is  then  represented  by 
AB,  and  so  forth. 

It  is  seen  that  in  this  case  the  resultant 
e.m.f.  is  less  than  the  algebraic  sum  of  the 
individual  e.m.fs.  of  the  coils.  It  is  the 
vector  sum  of  the  e.m.fs  and  is  2/7r  times 
the  algebraic  sum. 

If  the  total  winding  has  N  turns,  the 
equivalent  number  of  turns  of  a  concen- 
trated winding  would  be  T  =  2/ir  N. 


282  ELECTRICAL  ENGINEERING 

If  instead  of  being  distributed  all  around  the  periphery  the 
winding  covered  an  arc  of,  say,  60°,  as  is  shown  in  Fig.  221,  the 
effectiveness  would  again,  by  a  similar  diagram,  be  found  to  be 
the  ratio  of  the  chord  to  the  arc.  Thus,  the  chord  is  evidently 

2  sin  30°  and  the  arc  -£-• 

/"  A\.     \ 

,.  fc  _  i-taW  _  8  f          g;sft 


_..  7T 

T 

and 

T  =  0.955AT  FIG.  221. 

In  general,  if  the  winding  covers  a  electrical  degrees, 

a 


360    ir 

Example. — A  completely  distributed  single-phase  winding  has 

a  =  180°. 

.'.  *  =  2- 

7T 

Three-phase  winding  uniformly  distributed.     In  this  case,  the 
winding  covers  60°.     Thus,  k  =  0.955. 


CHAPTER  XXXVIII 
HUNTING 

The  periodic  oscillation  of  synchronous  machinery  is  a  familiar 
and  oftentimes  troublesome  phenomenon,  It  manifests  itself 
principally  by  the  swinging  of  the  needles  of  meters  connected  in 
the  circuits.  When  the  effect  is  cumulative,  it  continues  to  in- 
crease until  rupture  occurs  somewhere  in  the  system.  Often  it 
is  not  cumulative,  and  resembles  simply  the  movement  of  any 
vibrating  body  such  as  a  pendulum. 

The  difficulty  of  visualizing  hunting  of  a  revolving  machine 
comes  from  the  fact  that  the  vibration  is  superposed  on  the  steady 
rotation  of  the  moving  part.  It  can  be  well  imagined  as  similar 
to  the  motion  in  space  of  a  pendulum  swinging  east  and  west  while 
at  the  same  time  the  earth,  on  which  the  pendulum  is  fixed,  is  in 
rotation. 

Hunting  of  electrical  machines  is  possible  because  the  position 
of  the  armature  core  in  the  field  structure  at  any  moment  is  deter- 
mined by  the  balance  of  mechanical  and  electromagnetic  forces. 
Assuming  the  mechanical  force  to  be  steady,  as  represented  by  the 
shaft  or  belt  in  connection  with  the  prime  mover  or  load,  the 
electromagnetic  force  is  variable  owing  to  the  highly  elastic 
property  of  the  magnetic  field.  Under  absolutely  steady  condi- 
tions there  would,  of  course,  be  no  hunting.  But  such  conditions 
do  not  exist,  and  any  variation  of  the  electromagnetic  forces 
results  in  a  change  of  speed  as  the  machine  re-establishes  the 
momentarily  lost  equilibrium.  Hunting,  or  oscillating,  is  thus 
started  and  continues  as  equilibrium  is  gradually  restored  in  the 
elastic  medium  of  the  field. 

The  mechanical  force  is  not  always  steady.  Steam  engines, 
and  especially  gas  engines,  are  subject  to  pulsation  of  driving 
torque.  This  may  appear  in  the  generator  in  the  form  of  forced 
electrical  vibrations,  especially  where  the  machines  are  directly 
connected.  When  the  generator  is  free  to  oscillate  in  response  to 
any  impulse,  it  does  so  at  a  definite  rate  called  its  natural  period 
in  distinction  to  a  forced  period. 

283 


284 


ELECTRICAL  ENGINEERING 


The  natural  period  of  a  pendulum  depends  on  its  length  and 
mass,  the  length  being  the  radius  of  gyration.  Similarly  the 
natural  period  of  an  armature  or  revolving  field  structure  depends 
on  its  mass  and  radius  of  gyration. 

To  find  the  natural  period  of  a  machine,  consider  the  motion  of 
a  stretched  spring  as  illustrated  in  Fig.  222.  The 
spring  suspends  a  weight,  and  its  motion  is  damped  by 
a  piston  working  in  a  dash  pot. 

Let  F  =  pulling  down  force  in  the  spring,  y  =  dis- 
placement of  the  weight.  Then,  ft  =  ay  =  tension  on 
spring,  where  a  =  number  of  pounds,  per  unit  length, 
of  the  downward  pull. 

If  the  friction  force  due  to  the  dash  pot  is  assumed 
proportional  to  the  velocity,  the  force  necessary  to 

overcome  friction  =  //  =  k  -T-     (The  power  required 

varies  as  the  square  of  the  velocity.) 

The  force  required  to  overcome  inertia  =  M  or,  where 


Weight 


Dash  Pot 


FIG.  222. 


M  =  mass  and  a  =  acceleration, 


or, 


.  . 
=  M 


=  ay 


M 


is  the  total  force  required  to  balance  those  acting  in  the  system. 

If  the  applied  force  is  removed,  or  if  F  = 
0,  the  equation  becomes, 


dt 


=  0. 


Applying   this   equation   to  an  alternator,     ^-^yy^r^nm^-^ 
the   condition  is   as  illustrated  in  Fig.  223. 
The  moment  equation  is 

Pp  =  I  —  _}-  0  - — _|_  aQ}  FIG.  223. 

where  Fp  =  the  applied  moment,  F  being  the  force  and  p  the  lever 
arm. 

7  =  moment  of  inertia, 

6  =  initial  angular  displacement, 

)3  =  moment  of  retarding  force  per  unit  angular  velocity, 

a  =  twisting  moment  per  unit  angular  displacement. 

dB 

-J7  =  angular  velocity, 


HUNTING  285 


-77  =  moment  of  angular  velocity. 

p  =  radius  at  which  the  force  is  applied. 
la  =  moment  of  angular  acceleration. 
This  is,  by  ordinary  mechanics, 
Md2s  d  ds 


When  the  force  is  released,  Fp  =  0,  and 


The  solution  of  this  differential  equation  is 
0  =  Ae*V  +  BtmJ, 

in  which  A,  Bt  mi  and  niz  are  to  be  determined  from  known 
conditions. 
Let 

dO 


Then, 

m2!  +  0m  +  a  =  0, 
arid 

2_L^  « 

mz  -\-  j  m  =  —  jp> 
whence, 


_ 

-     ~  21  ± 


-  2l          4P  ~  I;  m2  =    "  27 
If  vp  —  j  is  positive,  then  5  is  real,  and  the  equation  shows  that 

6  gradually  decreases  to  zero  without  oscillation.     If,  however, 

j32         a 
jp  —  -j  is  negative  then  the  square  root  is  imaginary  and  6 

reaches  zero  after  a  certain  number  of  oscillations. 

02         a 
Thus,  hunting  can  take  place  only  when  ^  —  j  is  negative. 

Let  then 


Then 


«.«_._£_. 

5  =  7~4/2 


/3 

mi  =  -        + 


286  ELECTRICAL  ENGINEERING 

and 

a 

m*  =  ~  27  ~  jd 
or  putting 


Wi  =  -  7 
m2  =   -  7  - 

.'.   0  =  Ae-^ejs 


When  t  =  T,  the  time  of  one  period, 

6T  =  27T 


Assume  the  case  of  suddenly  throwing  off  full-load  from  the 
alternator.     Then  6  =  00. 

At  t  =  0,  the  hunting  has  not  yet  begun. 

" 


The  period, 

T=^=     ,M        , 
^        VW  -  |82 

where  /3  is  the  friction  torque,  and  has  little  influence  on  the  period 
of  hunting,  but  rather  affects  the  amplitude. 

We  may  assume  0  =  0. 
Then 


where  T7  is  in  seconds. 

The  beats,  or  oscillations  per  second,  are  ™>  or 


oo 
Beats  per  minute  =  —  * 

7T        \  x 


The  angular  space  position  of  the  alternator  armature  with 
reference  to  the  field  pole  may  be  determined  for  any  load  (Fig. 
224).  Let  this  angle  be  assumed  to  be  20°  for  a  two-pole  machine 


HUNTING  287 

at  full-load,  or  10°  for  a  four-pole  machine.     If  6  =  mechanical 
angle,  and  <f>  =  electrical  angle, 

8  =  —  where  v  = 
P 

number  of  poles. 
Torque, 

7050  X  kw.  . 

r.p.m. 

If    a   =   torque    per    unit 
angular  displacement, 

T 


57.3 

where  00  =  angle  in  degrees  for  the  load  being  considered. 
Therefore, 

=  24.25  X  kw.  X  /  X  10s 

where  /  =  frequency,  N  =  revolutions  per  minute  and  <t>  is  in 
degrees. 

Finally,  the  solution  is, 

Beats  per  minute,  Sm  = 

The  number  of  beats  per  minute  may  be  changed  by  changing 
/  or  </>°,  the  former  by  the  addition  of  a  fly-wheel,  the  latter  by 
altering  the  gap.  Bridges,  or  dampers,  between  the  poles  may 
also  be  used  to  produce  eddy  currents  for  the  purpose  of  damping 
the  oscillations. 

Problem  96. — Determine  the  periods  of  the  100-kw.  alternator  of  the 
previous  problems,  both  as  definite  pole  and  as  a  round  rotor  machine,  and 
with  long  and  short  gaps. 

Solution. — The  equation  is 

=  4^000      /kw.  X/ 

in  which  the  constants  previously  given  are. 

N  =  r.p.m.  =  900;  kw.  =  100;  /  =  60; 

TPr2      800*  X  0.862 
1  =  moment  of  inertia,  = =  ~     oo  i  A 

p  =  0.86  ft.  =  radius  of  gyration. 


288 


ELECTRICAL  ENGINEERING 


Supplying  numerical  values, 

47,000      /100  X60 
\18.4X  d>° 


942  \  -o 


tan  /3  =  tan  (5  — 


900      \18.4X<^ 
To  find  0°,  in  Fig.  224,  <f>  =  0  -  90°  +  a. 
Assuming  non-inductive  load, 

Ei  =  e  +  Ir  +  jlx  =  a  +  jb. 
F/  =  —be  —  ml  -f-  ,/aC  =  d  +  jf. 
f_  _b 

tan  5  —  tan  a     _  d       a        af  —  bd 
''  1  +  tan  5  tan  a  =          bf_  =  ad  +  bf 

+  ad 
where 

a  =  e  +  Ir  =  1330  +  25  X  0.69  =  1347.25, 

d  =  -bC  -  ml  =  -25zC-  25m, 
/  =  a<7  =  1347.25C. 

Tabulating,  for  the  four  cases : 


Definite  pole 

Round  rotor 

Gap,  in. 

0.25 

0.1875 

0.25 

0.1875 

X 

8.15 

14.1 

8.15 

14.1 

C 

2.75 

2.18 

2.75 

2.18 

m 

47,5 

47.5 

59.4 

59.4 

f 

3,700 

2,940 

3,700 

2,940 

b 

204 

353 

204 

353 

d 

-1,750 

-1,960 

-2,040 

-2,250 

of 

4,980,000 

3,960,000 

4,980,000 

3,960,000 

bd 

-357,000 

-692,000 

-416,000 

-795,000 

ad 

-2,360,000 

-2,640,000 

-2,745,000 

-3,030,000 

bf 

755,000 

1,040,000 

755,000 

1,040,000 

af  -  bd 

5,337,000 

4,652,000 

5,391,000 

4,755,000 

ad  +  bf 

-1,605,000 

-1,600,000 

-1,990,000 

-1,990,000 

tan/9 

-3.32 

-2.92 

-2.71 

-2.39 

/3 

106.75 

109° 

110.25 

112.7° 

tan  a 

0.1514 

0.262 

0.1514 

0.262 

a 

8.6° 

14.67° 

8,6° 

14.67° 

ft  -  90°  +  a 

25.35° 

33.67° 

28.85° 

37.37° 

1/0° 

0.0394 

0.0297 

0.0346 

0.0268 

Vl/*° 

0.190 

0.172 

0.186 

0.1635 

sm 

187 

162 

175 

154 

CHAPTER  XXXIX 


STUDY  OF  THE  DESIGN  CONSTANTS  OF  ALTERNATORS 

Alternators  differ  primarily  in  respect  to  the  number  of  phases, 
and  whether  the  armature  or  the  field  structure  is  the  revolving 
part.  Secondarily,  they  differ  in  respect  to  the  frequency, 
voltage,  output  rating  and  speed. 

In  practice,  the  very  great  majority  of  alternators  are  of  the 
three-phase,  revolving-field  type.  In  frequency,  they  are  gen- 
erally of  either  the  25-cycle  or  60-cycle  type  in  America;  25-  and 
50-cycle  in  Europe.  Voltage  may  be  any  desired  value  up  to 
about  13,000.  In  output  rating  alternators  are  built  up  to 
30,000  kva. 

The  speed  is  limited  by  the  prime  mover  and  the  frequency. 
Maximum  speed,  for  60-cycle  machines  is  3600  r.p.m.,  corre- 
sponding to  the  requirement  of  a  bipolar  field;  for  25-cycles,  the 
maximum  speed  is  1500  r.p.m.  The  chief  types  of  prime  mover 
used  with  alternators  are  the  reciprocating  engine,  representing 
moderate  speeds,  the  water  turbine  representing  low  speeds,"  and 
the  steam  turbine  representing  high  speeds.  Certain  roughly 
approximate  constants  have  been  obtained  from  experience 
which  may  serve  as  guides  in  preliminary  design.  These  are 
given  in  Table  IX. 

TABLE  IX. — APPROXIMATE  CONSTANTS  OBTAINED  FROM  EXPERIENCE 


Prime  mover 

Recip.  engine 

Water  turbine 

Steam  turbine 

Frequency.  .  . 

25 
5 
3,200 

2.5 

6 
2.5 

60 
3 
1,800 

2.5 

6 
2.5 

25 
13 
8,500 

2.5 

6 
2.5 

60 
5 
3,200 

2.5 

6 
2.5 

25 
20 
13,000 

2.5 

6 
2.5 

60 
7.5 

4,800 

2.5 

6 
2.5 

Arm.  dia.  per  pole  
Arm.  reac.  per  pole  
No  load  A.T.-per  pole 

Arm.  reac. 
Regulation  (approx.),  per 
cent  
Sh.  cir.  cur.  at  0  load  exc. 

Full-load  current 

19 


289 


290  ELECTRICAL  ENGINEERING 

Using  them  as  a  basis,  the  design  constants  will  be  calculated 
for  the  following  alternator: 

A.r.B-8-100-900-2300  volt. 

General  Constants.  —  From  the  rating  it  is  seen  that  the  machine 
is  a  three-phase,  revolving-field,  8-pole,  100-kilo  volt-amp.,  900- 
r.p.m.,  2300-volt  alternator,  evidently  to  be  driven  by  a  recipro- 
cating engine. 

It  is  first  necessary  to  decide  whether  the  phase  windings  shall 
be  connected  Y  or  A.  Y-connection  is,  in  general,  suitable  for 
higher  voltages  and  lower  currents.  Therefore  Y-connection 
will  be  assumed  in  this  case.  The  phase  winding  voltage  is  then 


The  phase  current  =  line  current 
Kva        100,000 


^  r.p.m.  ^,  poles       900  v 

Frequency  =     ^Q     X  —  ^—  =  -QQ   X  4  =  60  cycles. 

Slot  Dimensions.  —  The  development  of  the  design  now  depends 
on  the  determination  of  size  and  number  of  slots  and  the  conduc- 
tors *in  the  slot. 

It  has  been  found  that  for  an  n-phase  machine,  armature  reac- 

tion per  pole  =  ~  IPt,  where  t  =  effective  turns  per  pole  and  phase, 

z 

and  Ip  is  the  maximum  value  of  the  current  in  the  windings. 
For  three-phase,  therefore,  by  Table  IX, 

1800  amp.-turns  =  1.5\/2  X  25£. 

.  '  .  t  =  34. 

This  number  serves  as  a  good  preliminary  value.  Actually,  28 
turns  per  pole  per  phase  were  chosen.  Conductors  per  pole  and 
phase  are  then  2  X  28  =  56.  The  number  of  slots  per  pole  per 
phase  depends  primarily  on  the  armature  circumference  and  the 
slot  pitch.  With  many  slots,  a  smoother  e.m.f.  wave  is  generally 
obtained.  The  number  of  slots  is  also  usually  greater  in  low 
voltage  machines,  where  the  requirements  of  higher  insulation  are 
not  so  severe.  Practically,  at  least  two  slots  per  pole  per  phase 
are  used. 


DESIGN  CONSTANTS  OF  ALTERNATORS        291 


From  Table  IX,  the  armature  diameter  per  pole  is  found  to  be 
3  in.  Hence  the  diameter  is  3  X  8  =  24  in.  and  the  circum- 
ference is  TT  X  24  =  75.5  in. 

The  slot  pitch  may  be  determined  for  different  numbers  of  slots 
per  pole  per  phase,  as  follows: 


Slots  per  pole  per  phase  .  . 

1 

O 

3 

4 

Slots  per  pole 

3 

6 

9 

12 

18 

Slots 

24 

48 

72 

144 

Ql/^j.    .-.;.*.  ^U      /   *"•"    \      •         :r./,Vioe; 

31  A 

blot  pitcn  1   ,   ,    i  in  incnes  

.  1* 

.57 

.05 

.  785 

0.524 

About  half  of  the  slot  pitch  will  be  required  for  the  tooth. 

Considering,  therefore,  the  insulation  requirement  given  in  Table 
X,  it  is  fairly  apparent  that  a  small  number  of  slots  per  pole  per 
phase  should  be  chosen.     It  will  be  assumed  that. 
there   are   2   slots  per  pole  per  phase.     Each  slot 
will  then  contain, 

56 

-j-  =  28  conductors. 


A  very  good  arrangement  of  conductors  in  a  slot 
is  that  shown  in  Fig.  225,  which  permits  of  easy 
insertion  of  the  coils. 

TABLE  X 


Voltage  (phase) 
110 

440 

,  1,000 

2,300 

6,600 

16,000 


Insulation,  a 

20  mils 
25 
35 
50 
90 
130 


The  size  of  the  conductor  must  next  be  determined.  As  a  guide 
for  this,  it  may  be  taken  as  permissible  to  use  current  densities 
up  to  2500  amp.  per  sq.  in.  in  low-voltage  machines,  and  up  to 
1200  amp.  per  sq.  in.  in  high-  voltage  machines.  Assuming  a 
density  of  2000  as  reasonable  the  required  area  of  conductor  to 
carry  25  amp.  is 


X 


It  is  seldom  good  practice  to  use  wire  heavier  than  No.  10  B.  &  S. 
As  0.0125  sq.  in.  corresponds  nearly  to  No.  8,  it  will  be  preferable 


292  ELECTRICAL  ENGINEERING 

to  divide  this  area  among  several  wires  in  parallel.  The  con- 
ductor used  consists  of  four  No.  14  wires  in  parallel,  having  a 
combined  cross-section  of  4  X  0.00323  =  0.01292  sq.  in.,  and 
giving  a  resultant  current  density  of  1925  amp.  per  sq.  in. 

The  arrangement  of  wires  in  the  slot  is  similar  to  that  of  Fig. 
225.  There  are  four  groups  of  28  wires  each,  the  wires  being 
placed  four  abreast  and  seven  deep.  Each  layer  of  four  wires  is 
insulated  from  those  above  and  below  it. 

Width  of  copper  in  the  slot  is  8  X  0.064  in.  =  0.512  in. 
Width  of  insulation  =  0.238  in. 
Width  of  slot  =  0.512  +  0.238  =  0.75  in. 
Depth  of  copper  in  slot  =  14  X  0.064  =  0.896 
Depth  of  insulation         =  0.59 
Depth  of  wedge  =0.2 

Depth  of  slot  =  1.686  =  1  *K6  in. 

Width  of  tooth  at  face  =  slot  pitch — slot  width  =  1.57  —  0.75 
=  0.82  in. 


circumference  at  base 

no.  teeth 
TT  X  (24  +  3.375) 


Width  of  tooth  at  base  =  -          ^  tee -th      '       "  °'75 


48  -  0.75  =  1.038  in. 

Flux  Determination. — The  general  equation  for  effective  e.m.f. 
per  phase  is 


108 
where 

4.44  =  4  X  6ffeCtive  e"mf  =  4  X  -^=  =  4  X  1.11 
average  e.m.f. 


4>a  —  total  flux  per  pole  entering  the  armature  at  no-load, 
t  =  total  armature  turns  in  series  per  phase,  =  8  X  28  =  224, 
/  =  frequency  =  60, 

k  =  constant  depending  on  the  distribution  of  conductors  on 
the  armature  periphery. 

If  the  conductors  were  concentrated  in  a  single  slot  per  pole 
per  phase,  k  would  be  1.  With  a  three-phase  machine,  these  con- 
ductors would  never  be  spread  out  over  the  entire  180  electrical 
space  degrees  of  the  pole  pitch  as  in  single-phase  or  direct-cur- 
rent machines,  but  would  be  restricted  to  one-third  of  this 
amount,  or  to  60°,  on  account  of  the  space  required  for  the  other 


DESIGN  CONSTANTS  OF  ALTERNATORS        293 

phases.     Where  there  are  two  slots  per  pole  per  phase  the  e.m.fs. 
generated  in  the  two  slots  add  vectorially,  as  illustrated  in  Fig. 

226,  where  E  =  EI  +  E2.     k  is  then  evidently  equal  to  ^-^ — • 
For  n  slots  per  pole  per  phase, 

1 


o      •    60° 
2n  sin  7:— 

2n 


Thus,  for 


Supplying  these  values  in  the  e.m.f.  equation  and  solving  for 
flux, 

2300  1 

*«  - 


V  0.966X60X224X4.44 

=  2.3  megalines. 

The  flux  leakage  factor  for  this  machine  is  1.125. 
.'.  flux  in  the  field  at  no-load  is, 

4>f  =  2.3  X  1.125  =  2.59  megalines. 

.  Air  Gap.  —  An  approximate  average  value  for  the  gap  length 
may  be  obtained  by  reference  to  Table  IX.     In  the  table  is  found, 

no-load  A.T.  per  pole  _ 
arm.  reaction 

Armature  reaction  =  1.5  X  A/2  X  25  X  28  =  1490. 
Substituting  this  value  of  armature  reaction, 

no-load  A.T.  per  pole  =  2.5  X  1490  =  3725. 

These  ampere-turns  are  mostly  required  for  the  gap.     Assum- 
ing 80  per  cent,  for  this, 

gap  A.T.  =  0.8  X  3725  =  2980. 

Assuming,  now,  a  gap  flux  density  at  no-load  of  40,000  lines 
per  sq.  in.,  and  substituting  in  the  equation, 

A.T.  (gap)  =  0.313  B0  X  la, 
where  lg  is  the  length  of  one  gap, 

2980  =  0.313  X  40,000  X  10, 


0.313  X  40,000 


=  °'238 


294  ELECTRICAL  ENGINEERING 

With  this  value  for  a  guide,  definite  values  may  be  chosen. 
With  alternators,  it  is  usual  to  shape  the  pole  pieces  so  that  the 
generated  e.m.f.  may  more  nearly  approach  the  sine  form.  The 
gaps  chosen  for  this  machine  are : 

gap  length  in  center  of  pole        =  0.1875  in. 

gap  length  at  edge  (maximum)  =  0.386    in. 

average  gap  length,  lg,  =  0.2535  in. 

Gap  area, 

flux  2.3  X  106       __  K 

A°  =  fluTdelisIty  =      40,000      =  57'5  Sq"  m' 

Armature  Length. — The   main   factor   bearing   on   armature 
length  is  flux  density  in  the  teeth.     This  in  turn  depends  upon 
gap  area,  pole  pitch  and  pole  arc. 
The  pole  pitch  at  the  armature  surface  is 


The  pole  arc  is  usually  about  0.6  X  pole  pitch. 
In  this  machine,  the  ratio 

P°le  arc     _  0*0 

i  '.L    l~    —    U.OO. 

pole  pitch 

Assuming  pole-face  area  =  air-gap  area,  length  of  pole  piece 
parallel  to  shaft  is 

A  A7  K 

=  11.5  in. 


Ag  57.5 


0.53  X  9.43  5 
The  armature  gross  length  may  be  slightly  greater  than  this  to 
assist  in  the  free  balancing  in  the  field.  The  gross  length  is  there- 
fore taken  as  12  in.  This  length  would  justify  the  use  of  four 
J^-in.  ventilating  ducts,  one  for  every  3  in.  The  length  of  lami- 
nations is  therefore  12  in.  —  2  in.  =  10  in.  Assuming  10  per 
cent,  loss  of  length  due  to  insulation  between  laminations,  the 
net  armature  length  is 

la  =  10  X  0.9  =  9  in. 
The  ratio, 

effective  length       _9_ 

total  length       =  12  = 

Teeth  Flux  Density. — Allowing  10  per  cent,  extra  for  "fring- 
ing" of  the  flux  entering  the  armature  from  the  pole  face,  the 
average  number 'of  teeth  under  the  pole  is 

fi^S  X  L1°  -  07  X  L1  "  3'5 


DESIGN  CONSTANTS  OF  ALTERNATORS        295 

This  number  varies  from  moment  to  moment  according  as  a 
slot  or  a  tooth  is  in  the  center  line  of  the  pole.  Teeth  area  at 
armature  face  is  then, 

At  =  3.5  X  tooth  width  X  effective  length  of  armature. 

In  order  that  this  area  shall  carry  a  flux  density  of  about  90,000 
lines  per  sq.  in.  in  the  tooth,1  it  may  be  calculated  on  that  basis. 
Thus,  the  flux  entering  the  armature  at  no-load  is  2.3  X  106  lines. 

2.3  X  106 
•'  At=      90,000      =  256  sq'  m" 

From  this  value  of  area,  the  length  obtained  is, 
256 


3.5  X  0.821 


8.9  in. 


Thus,  the  length  of  9  in.  previously  obtained  is  quite  satisfac- 
tory, giving,  as  it  does,  a  slightly  less  teeth  density  at  no-load, 
but,  as  will  be  seen,  approximately  90,000  at  full  non-inductive 
load. 

Armature  Resistance.  —  All  data  has  now  been  obtained  that  is 
necessary  to  calculate  the  resistance  of  the  armature  winding. 
The  length  of  the  mean  turn  may  be  taken  as  twice  the  gross 
length  of  the  armature  core  plus  nine  times  the  diameter  per 
pole;  or  the  length  of  the  mean  turn 

=  2L  +  9  D/pole 

=  2  X  12  +  9  X  3  =  51  in. 

-g-  4.25  ft. 

Total  length  =  turns  per  phase  X  mean  turn. 

=  8  X  28  X  4.25  =  954  ft. 

2  9 
Resistance  of  four  No.  14  wires  in  parallel  =  -j-  =  0.725  ohms 

per  1000  ft. 

/.  Ra  per  phase  =  X  0.725  =  0.69  ohm  at  60°C. 


For  25-cycle  alternators  110,000  lines  per  sq.  in.  is  suitable. 


296 


ELECTRICAL  ENGINEERING 


Voltage  drop  per  phase  =  IRa  =  25  X  0.69  =  17.25  volts. 
The  full-load  e.m.f.  per  phase  =  E  +  IRa  approximately. 

=  1330  H-  17.25  =  1347  volts 

Magnetic  Circuit  Dimensions. — Sufficient  data  is  now  at  hand 
to  enable  the  making  of  a  sketch  which  shall  show  approximately 
how  the  available  space  may  be  utilized. 

Fig.  227  represents  such  a  section  of  the  magnetic  circuit.  The 
next  step  is  to  construct  a  table  for  the  condition  of  no-load  and 
normal  voltage,  from  which  is  obtained  the  total  required  number 
of  field  ampere-turns.  Some  of  the  data  in  this  table  have  already 
been  obtained,  especially  the  required  fluxes  in  the  different  parts. 


FIG.  227. 


The  yoke  is  left  out  of  consideration,  its  magnetic  length  being 
very  small  in  revolving  field  machines  of  few  poles. 

The  armature  and  pole  sectional  areas  are  arbitrarily  chosen  to 
give  appropriate  densities.  The  length  of  the  pole  depends  upon 
the  space  required  by  the  field  winding.  The  field  values  ob- 
tained for  this  machine  are  given  in  Table  XI. 

Material  of  the  armature  core  is  standard  sheet  iron  of  0.014  in. 
thickness.  The  field  core  is  built  up  of  thick  steel  punchings. 


DESIGN  CONSTANTS  OF  ALTERNATORS        297 
TABLE  XI 

Magnetic  data.     No-load,  normal  voltage 


Part 

Flux  (mgl.) 

Area 

B 

A.T.  per  in. 

Length 

A.T. 

Teeth  

2.3    (face) 
(base) 
2.3 
1.15 
2.59 

26.0 

32.8 
57.5 
28.2 
27.5 

89,000 
70,000 
40,000 
41,000 
94,200 

15.  6\ 

7.0/11'3 

2.8 
74.2 

1.6875 

0.2535 
6.0 
6.25 

19 

3,150 
17 
464 

3,650 

21 

3,203 
17 

484 

Gap 

Arm  

Pole  

Total  amp.  -turns  . 
Teeth 

Full 

2.  34  (face) 
(base) 
2.34 
1.17 
2.64 

-load,  i 

lormal  v 
90,500 
71,200 
40,700 
41,700 
96,000 

oltage 

^o}"-' 

Gao 

Arm 

2.85 

77.5 

Pole  

Total  amp.  -turns  . 

3,725 

To  the  total  required  ampere-turns  to 
excite  the  field  at  full-load  must  be 
added  those  necessary  to  compensate  for 
the  armature  reaction.  The  number 
3725  is  the  resultant,  Fr,  Fig.  228.  The 
total  ampere-turns  on  the  field  core,  F/, 
must  be  equal  to  the  vector  difference  of 
Fr  and  Fa,  where  Fa  is  the  armature  am- 
pere-turns multiplied  by  the  field  leakage 
factor.  For  full  non-inductive  load,  ap- 
proximately 


1.125  r 


FIG.  228. 


Ff  = 
Supplying  values  already  obtained, 

Ff  =  \&7252  +  L125~>04902  =  4090  A.T. 

For  any  other  power  factor,  say  80  per  cent.,  the  required  field 
ampere-turns  are  approximated  as  illustrated  by  dotted  lines  in 
Fig.  228.  Thus, 

F'f  =  MFr  +  1.125fl.  sin  02  +  l.l25Fa  cos  02 

=  \3725  +  1677lxf<X62  +  1677  X  0.82  =  "N/47322  +  13412 
=  4920  A.T. 


298  ELECTRICAL  ENGINEERING 

The  Main  Field  Magnetomotive  Force.  —  The  ampere-turns 
which  must  be  supplied  to  each  field  pole  are: 

for  no-load,  normal  voltage,  3650 
full-load,  non-inductive,  4090 
full-load,  80  per  cent,  lagging,  4920 
maximum  exciter  voltage  =  110. 

The  field  winding  may  be  taken  as  composed  of  copper  strip, 
edge  wound.  The  depth  of  such  a  winding  may  vary  from  %  in. 
to  1  Y±  in.  under  ordinary  circumstances,  being  usually  deeper  with 
short  poles. 

The  choice  of  the  actual  dimensions  for  a  given  case  is  largely 
a  matter  of  experience.  The  limiting  factor  is,  of  course,  the 
amount  of  heat  that  may  be  radiated. 

In  this  machine,  the  field  conductor  is  0.625  in.  wide  by  0.0175 
in.  thick. 

Length  of  winding  space,  exclusive  of  that  required  for  pole 
insulation  =  5.5  in. 

Turns  in  series  per  spool  =  230.5 

3650 
Field  current,  no-load       =  oon  -  =  15.8  amp. 


Field  current,  full-load,  non-inductive  =  17.8  amp. 

Field  current,  full-load,  80  per  cent,  lagging  =  21.3  amp. 

Mean  length  of  field  turn  =  2.72  ft. 

Total  length  of   field   winding  (8  spools)  =  8  X  230.5  X  2.72 

=  5020  ft. 

Cross-section  of  conductor  =  0.01095  sq.  in. 
Resistance,  at  60°C.  =  4.3  ohms. 
Excitation  volts,  no-load  =  15.8  X  4.3  =  67.5 
Excitation  volts,  full-load,  non-inductive  =  76.5 
Excitation  volts,  full-load,  80  per  cent,  lagging  =  91.5 
Current  density  in  the  field  winding: 

no-load  =  1434  amp.  per  sq.  in. 
full-load,  non-inductive  =  1625 
full-load,  80  per  cent,  lagging  =  1945 

Losses  and  Efficiency.  —  Full-load,  non-inductive. 
Armature  copper  loss  per  phase  =  PRa  =  252  X  0.69  =  432. 
Total  copper  loss  in  armature  =  3  X  432  =  1296  watts. 
Field  copper  loss  =  J/2#/  =  17.82  X  4.3  =  1370  watts. 


DESIGN  CONSTANTS  OF  ALTERNATORS        299 

The  core  losses  have  already  been  calculated  for  direct-current 
machines.  The  hysteresis  loss  for  alternators  is  determined  on 
the  same  basis.  The  eddy  current  loss  will  not,  as  in  direct- 
current  machines,  be  equal  to  the  hysteresis,  but  owing  to  the 
greater  degree  of  lamination,  will  be  much  less. 

In  this  case  it  will  be  assumed  that  the  eddy  current  loss  is  50 
per  cent,  of  the  hysteresis  loss. 

Weight  of  armature  core  =    806  Ib. 

Weight  of  teeth  =    180  Ib. 

Hysteresis  loss  in  core  =  1130  watts 

Hysteresis  loss  in  teeth  =    730  watts 


Total  hysteresis  loss  =  1800  watts 

Total  iron  loss  =  1.5  X  I860  =  2790  watts 

Friction  and  windage  loss,  assumed  1  per  cent.  =  1000  watts 
Total  loss,  full-load,  non-inductive  =  6460  watts 


Efficiency  =  =  0.939. 


Temperature  Rise.  —  This  is  determined  for  the  different  parts 
by  the  use  of  coefficients  obtained  in  practice.  For  rotating 
armature  machines  the  radiation  of  0.8  watt  per  sq.  in.  of  surface 
corresponds  approximately  to  a  temperature  rise  of  100°C. 
Thus,  for  a  rise  of  40°,  the  radiating  surface  should  be  sufficient 
to  dissipate  0.3  watt  per  sq.  in. 

With  rotating  field  cores,  owing  to  the  greater  fanning  action 
a  larger  amount  of  energy  is  dissipated  for  the  same  temperature 
rise.  In  some  cases,  particularly  with  turbo-alternators,  there 
are  placed  on  the  revolving  structure  fan  blades  which  increase 
the  heat  dissipation  still  more.  In  such  cases,  2  watts  per  sq.  in. 
might  correspond  to  a  100°  temperature  rise.  The  actual  tem- 
peratures which  different  parts  of  a  given  machine  will  attain 
can  only  be  estimated  from  experience,  from  the  current  and 
flux  densities  and  from  a  study  of  the  particular  structure  with 
relation  to  the  ventilating  action  which  it  will  produce. 

In  the  machine  under  consideration,  the  copper  loss  in  each 

field  winding  is  ~~o~  =  171  watts. 

The  area  of  the  coil  surface,  including  both  the  external  and 
the  internal  surfaces,  is  about  400  sq.  in. 

Watts  per  sq.  in.  radiated  are  therefore  J      =  0.427. 


300  ELECTRICAL  ENGINEERING 

It  is  safe  to  assume  that  this  will  not  cause  a  temperature  rise 
greater  than  40°C.  The  total  loss  in  the  armature  is  4090  watts. 
This  is  dissipated  from  a  total  area,  including  air  ducts,  and  al- 
lowing for  extension  of  the  end-connections,  of  about  8200  sq.  in. 

4090 
Watts  per  sq.  in.  radiated  are  therefore,  QOAA  =  0.5,  which  is 


entirely  conservative. 

Regulation.  —  This  may  be  determined  directly  from  the  satura- 
tion (magnetization)  curve.  Thus,  the  field  excitation  for  full 
non-inductive  load  has  been  found  to  be  4090  amp.-turns.  Re- 
ferring to  the  curve  for  this  machine,  shown  in  Fig.  210,  the  no- 
load  voltage  with  this  excitation  is  2440. 

2440  —  2300 
.'.  Regulation  =  -      OQnA      -  =  0.061  =  6.1  per  cent. 


Regulation  may  also  be  determined  by  adding,  vectorially, 
the  IR  and  IX  drops  to  the  full-load  voltage,  to  obtain  the  no- 
load  voltage.  The  reactance  has  been  seen  to  have  two  com- 
ponent values  representing  that  of  the  coils  immediately  under 
the  poles,  and  that  of  the  coils  between  the  poles.  These  have 
been  designated  x  and  Xi,  respectively,  and  their  values  for  this 
particular  machine  have  been  determined  in  connection  with 
problem  90.  The  theoretical  determination  of  x  has  also  been 
carried  out  in  connection  with  Fig.  210. 


CHAPTER  XL 
SHORT-CIRCUIT  OF  ALTERNATORS 

The  short-circuiting  of  a  direct-current  generator  is  a  very 
serious  event.  The  commutator  usually  "flashes  over"  and  the 
belts  or  shafts  are  dangerously  strained.  With  alternators,  ex- 
cept with  large  turbo-machines,  such  short-circuit  results  in 
practically  no  excessive  stresses  and  of  course  not  in  any  "fire- 
works." 

However,  the  phenomena  of  alternator  short-circuits  are  of 
great  interest  and  importance.  They  involve  the  passage  from  one 
steady  state  —  that  of  normal  operation  —  to  another  steady  state  — 
that  of  the  permanent  short-circuited  condition.  Between  these 
two  steady  states  is  what  is  called  the  transient  period,  during 
which  the  system  is  thrown  out  of  equilibrium.  It  is  during  the 
transient  period,  especially  its  first  p^rt,  that  difficulties  some- 
times occur,  and  consequently  the  interest  of  the  student  lies 
chiefly  here. 

In  any  circuit  of  resistance  and  inductance,  in  which  a  con- 
stant e.m.f.  is  acting,  the  current  flowing  at  any  instant  after  the 
closing  of  the  switch  has  a  value  expressed  by  the  equation 

i  =  7(1  -€~£'). 

Similarly,  when  the  e.m.f.  is  removed  and  the  circuit  is  closed 
upon  itself,  the  current,  and  therefore  also  the  flux,  dies  down 
according  to  the  equation 


According  to  this  latter  equation  the  effect  of  resistance  is  to 
damp  out  the  current,  while  the  inductance  tends  to  maintain 
it.  A  most  striking  illustration  of  these  effects  is  afforded  by  the 
experiment  of  ONNEs,1  who  withdrew  a  magnet  from  a  closed  coil 
immersed  in  liquid  helium.  The  temperature  of  the  coil  was  so 
low  that  the  resistance  became  a  negligible  quantity,  and  the 
current  continued  to  flow  for  hours. 

Communication  No.  119  from  the  Physics  Laboratory,  Leiden. 

301 


302  ELECTRICAL  ENGINEERING 

Applying  these  equations  to  an  armature  under  the  condition 
of  short-circuit,  the  current  could  be  found  from  known  values 
of  r  and  L,  were  it  not  for  the  e.m.f.  of  rotation  of  the  armature 
in  the  resultant  field.  To  find  the  current  under  actual  condi- 
tions requires  first  a  knowledge  of  the  flux  at  any  instant,  and 
then  the  derivation  of  the  electromotive  force  from  the  flux. 

Thus,  at  the  instant  of  short-circuit,  it  may  be  assumed  that 
the  alternator  has  its  full  field  flux.  After  the  permanent  short- 
circuited  condition  has  been  reached,  the  field  has  fallen  to  only 
a  few  per  cent,  of  its  normal  value  on  account  of  the  armature 
reaction  (that  is,  the  armature  reactive  magnetomotive  force), 
which  demagnetizes  the  field.  During  the  transient  period  the 
field  is  not  much  affected  by  the  fluctuations  of  armature  current, 
these  being  balanced  if  the  field-circuit  resistance  is  low,  as  is 
always  the  case,  by  mutual  induction  with  the  field  circuit,  the 
field  and  armature  ampere-turns  acting  in  opposition  to  each 
other.  At  the  instant  of  short-circuit,  therefore,  the  value  of 
the  current  produced  depends  almost  entirely  upon  the  resistance, 
r,  and  the  reactance,  x,  of  the  armature.  Armature  reaction,  or 
the  demagnetizing  effect  of  the  armature  current,  has  no  appre- 
ciable effect,  at  first,  in  cutting  down  the  resultant  flux.  The 
current  may  rise  to,  say,  20  times  its  normal  value.  To  main- 
tain this  current  would  require  an  abnormally  large  field  excita- 
tion, many,  many  times  as  great,  in  fact,  as  that  which  actually 
is  available.  Indeed,  it  might,  without  great  error,  be  assumed 
that  in  comparison  the  actual  excitation  is  practically  zero.  In 
that  case,  then,  the  main  field  flux  is  surrounded  by  a  short-cir- 
cuited winding,  and  it  must  therefore  decrease  in  value  according 
to  some  exponential  law,  such  as, 


or,  if  time  is  expressed  in  radians 


The  final  value  of  the  flux  is  determined  by  what  is  known  as 
the  synchronous  impedance  of  -the  armature  which  consists  of  its 
resistance  and  the  equivalent  reactance  of  the  armature  magneto- 
motive force.  This  fact  fixes  the  values  of  r0  and  XQ. 

The  ratio  —  is  not  readily  calculated.     It  depends  not  only 

XQ 


SHORT-CIRCUIT  OF  ALTERNATORS  303 

upon  almost  all  constants  of  the  generator,  such  as  the  armature 
reaction,  armature  resistance,  field-circuit  resistance,  field  wind- 
ing, eddy  currents  in  field  poles,  etc.,  but  also  upon  the  nature 
of  the  short-circuit,  whether  single-phase  or  multiphase.  Suffice 
it,  therefore,  in  this  elementary  treatise,  to  state  the  fact  that  in 
almost  all  types  of  machines  it  is  in  the  neighborhood  of  from 
0.01  to  0.02.  In  other  words,  the  field  flux  dies  down  very 
slowly,  requiring  several  cycles  before  it  reaches  a  small  value. 
Since  the  speed,  during  the  transient  period,  may  be  assumed 
uniform,  the  induced  e.m.f.  will  decrease  according  to  the  same 
exponential  as  governs  the  flux. 
If  the  initial  value  of  the  e.m.f.  is 

c  =  Em  sin  ut 
and  the  final  value  is 

e2  =  Ezm  sin  cot, 
then,  during  the  transient  period,  the  e.m.f.  is 

e  =  Eime~  Lo    sin  ut  -f-  E2m  sin  at, 

that  is,  it  is  the  sum  of  the  final  value  and  a  transient  term,  the 
latter  being  proportional  to  the  instantaneous  value  of  the  flux. 
Re- writing  this  equation  in  terms  of  the  phase  angle,  0, 

--(0-00   ,  m?^ 

e  =  Eime     xo          sin  0  +  E2m  sin  0 

in  which  0i  is  the  phase  angle  at  the  instant  when  short-circuit 
occurs. 

0  —  01  represents  any  time  elapsing  after  the  instant  of  short- 
circuit. 

At  the  moment  of  closing  the  switch, 

Em  sin  0  =  Eim  sin  0  +  EZm  sin  0, 
and 

77T  T7T  |          77T 

&m    —    &lm  ~T~    &2m- 

Since  the  electromotive  force  in  (117)  acts  through  the  armature 
circuit  of  resistance,  r,  and  reactance,  x,  the  fundamental  Eq. 
(15)  will  evidently  hold,  and, 


— <8  —  0 } 
e  =  Eim  €     *<>  sin  0  +  EZm  sin  0  =  ir  +  x  - 


304  ELECTRICAL  ENGINEERING 

The  solution  of  this  is 

-  T—8   r     r    T—9   Elm     -  —  (0-0i)       . 

t  =   e     *     \  J  e.x        —  e     xo  sin  6dO 

l_  4-C 


sn 


_L0  rElm  -r~0ffl  r  £0   . 
=  e     x          —exojex    si 
l_   x 

Ezm    r  r-0     .        ,  j/l     i      /-» 

+  -    -J  *x  sm  0  rf(9   +  Ct 

.C  J 


where 

R       fr       r0\ 

^  =  (  ---  )  =  cot  |8. 

X        \3        Xo/ 

Substituting 

Z2  =  .R2  +  X2, 
and 

tan  /3i  =  -» 

and  determining  C  from  the  condition  that  when  0  =  0i,  i  =  o 
the  final  solution  is  given  by 


= 


This  equation  may  be  greatly  simplified  by  introducing  certain 
approximations,  which,  for  practical  considerations,  do  not  injure 
the  value  of  the  results  obtained.  Thus,  in  practice  E^m  lies 
between  2  per  cent,  and  10  per  cent,  of  Em,  .being  smaller  in 
larger  machines. 

Neglecting  E2m  in  (118),  and  writing  Em  for  Elm,  (118)  be- 
comes 

i  =  ^  ^ [€-r^(e  ~  9l)  sin  (B  -  0)  -  e~r*(0  ~  &l}  sin  (0!  -  0)] 

(119) 

Equation  (119)  is  convenient  for  fairly  accurate  work  and  should 
be  used  for  ordinary  wave  determinations.  Nevertheless,  rough 
approximations  may  be  made  by  further  simplification.  Thus, 

assume   0  =  90°;   then   sin    (0  —  0)  =  —  cos  0.     Also,   assume 

Y 

w  =  1.     Then  (119)  becomes 


SHORT-CIRCUIT  OF  ALTERNATORS  305 

These  assumptions  are  more  or  less  reasonable  since,  in  practice, 
/3  lies  between  85°  and  90°,  and,  in  concentrated  field  windings, 
the  reactance  is  much  greater  than  the  resistance. 

The  condition  for  maximum  current  is  when  61  =  o,  and  6  =  TT. 
Then, 


Em     -r-* 


. 
+e 


L«,-i 
xo  J. 


The  value  of  —  is  about  0.02  in  all  alternators,  and  ^r  =  0.06. 

XQ  XQ 

giving  e~0-06  =  1  approximately. 
Therefore  the  maximum  current  at  short-circuit  is 

Em     -JL. 


Continuing  the  evaluation,  -  is  from  0.6  to  0.8. 

x 
jp 

.'•    We  =  -  -  X  1.75  (approximately). 
x 

As  an  example,  take  an  alternator  which  has  4  per  cent,  react- 
ance. The  greatest  possible  current  that  can  be  obtained  on 
short-circuit  is  then 

imax  =  Q-QJ_  X  1.75  =  44  times  normal  current. 

To  illustrate  the  effects  of  short-circuit,  three  typical  generators 
are  taken  as  examples,  as  follows: 

Class  A.  —  Engine  driven  generators.  Reactance,  x  =  12  per 
cent.,  =  0.12,  resistance,  r  =  1  per  cent.,  =  0.01,  short-circuit 
current  under  normal  no-load  excitation,  Is  =  27,  where  /  = 
full-load  current. 

Class  #.—  Turbo-generators.    x  =  0.02,  r  =  0.01,  /,  =  21. 

Class  C.  —  Turbo-generators  with  external  reactance,  x  = 
0.06  (0.02  internal,  0.04  external),  r  =  0.01,  I8  =  21. 

All  three  machines  are  taken  on  the  percentage  basis;  with 
Em  =  1,  Im  =  1.  All  are  single-phase  generators,  or,  the  short- 
circuit  may  be  regarded  as  that  of  one  phase  only,  of  a  multiphase 
generator. 

Problem  96.  —  From  the  above  data  calculate  and  plot  the  first  few 
cycles  (2  to  4)  of  armature  current,  voltage  and  power. 

The  current  may  be  determined  from  (119),  the  voltage  from 
(117)  in  which  Ezm  is  neglected,  and  the  power  from  the  funda- 
mental relation,  p  =  ei,  where  instantaneous  values  are  con- 
sidered. 

20 


306 


ELECTRICAL  ENGINEERING 


The  following  values  are  at  once  obtained: 
r0  is  taken  equal  to  r;  XQ  =  j-  =  ^ 

18  £1 


~  =  0.5;  --  =  0.02 

6  XQ 


Class  A 

Class  B 

Class  C 

o                         /~«       *.  \    r 
K               .            IT        '0\ 
-^r=    COt  /3=   (  )    j 

0.0833  -  0.02 
0.0633 

0.5  -  0.02 

0.48 

0.1667  -  0.02 
0.1467 

2    =8^)3  = 

0.998 

0.9013 

0.9894 

0  = 

86°  20' 

64°  20' 

81°  40' 

The  only  other  constant  factor  remaining  to  be  supplied  is  0i, 
the  time-phase  angle  representing  the  instant  of  closing  the 
switch.  61  may  be  taken  at  any  desired  value,  and  it  should  be 
considered  what  effects  are  produced  with  different  values.  For 
convenience  of  calculation,  and  also  to  work  under  extreme  con- 
ditions the  following  values  of  0i  may  be  chosen. 


Class  A 

Class  B 

Class  C 

01   =    -  3°        40' 

-25°     40' 

-8°       20' 

86°     20' 

64°     20' 

81°     40' 

41°     20' 

19°     20' 

36°     40' 

For  each  value  of  0i  a  set  of  three  curves  may  be  obtained,  and 
a  comparative  study  will  then  be  possible,  both  in  regard  to  the 
effect  of  closing  the  switch  at  a  different  point  in  the  cycle  and 
with  regard  to  the  influence  of  the  constants  of  the  different  types 
of  machine.  In  the  present  instance  the  curves  for  the  engine 
driven  generator  (class  A)  are  produced  under  the  condition  0i  = 
—  3°  40'.  The  equations,  with  numerical  values  supplied,  are: 

i  =  8.32[e-°-02('  +  0-064)  sin  (0  -  86°  20')  +  €-°-0833C  +  °-064> 

=  8.32[e~a:  sin  a  +  €~v]  =  8.32[a  +  6] 
e  =6-o.02(*  + 0.064)  gin  0  =  6-*sin0 

p  =  ei  =  8.32[e-°-04(* +  0'064)  (sin2  0  cos  86°  20' 

-  sin  0  cos  0  sin  86°  20')  +  €-°-1033^  +°-064)  sin  0]. 

It  is  not  necessary  to  evaluate  the  power  equation  since  the 
product  ei  may  be  taken  for  each  angular  position.  The  tabula- 
tion is  given  for  360°  from  the  instant  of  closing  the  switch.  The 
three  curves  are  shown  in  Fig.  229  for  something  over  two  cycles. 
Figs.  230-237  are  for  the  other  cases  which  have  been  taken. 


SHORT-CIRCUIT  OF  ALTERNATORS 

Tabulating:  Case  A.     0t  =  -  3°  40'. 


307 


0-01 

0 

15 

30 

45 

60 

75 

90 

120 

0° 

-3°  40' 

11°  20' 

26°  20' 

41°  20' 

56°  20' 

71°  20' 

86°  20' 

116°  20' 

sin  0 

-0.064 

0.1965 

0.4436 

0.6604 

0.8323 

0.9474 

0.998 

0.8962 

a  =  0-86°20' 

-90° 

-75° 

-60° 

-45° 

-30° 

-15° 

0.0 

30° 

sin  a 

-1.0 

-0.9659 

-0.866 

-0.707 

-0.5 

-0.2588 

0.0 

0.5 

0(rad.) 

-0.064 

0.198 

0.459 

0.721 

0.982 

1.244 

1.507 

2.03 

0  +  0.064.... 

0.0 

0.262 

0.523 

0.785 

1.046 

1.308 

1.571 

2.094 

X 

0.0 

0.00524 

0.01046 

0.0157 

0.02092 

0.02615 

0.03142 

0.04188 

rx 

1.0 

0.9947 

0.9895 

0.984 

0.979 

0.974 

0.969 

0.959 

y 

0.0 

0.0218 

0.0436 

0.0654 

0.0872 

0.109 

0.131 

0.1745 

b  =  e-» 

1.0 

0.978 

0.957 

0.936 

0.916 

0.898 

0.878 

0.841 

a 

-1.0 

-0.96 

-0.857 

-0.695 

-0.49 

-0.252 

0.0 

0.48 

a  +  6 

O.Q 

0.018 

0.1 

0.241 

0  .  426 

0.646 

0.878 

1.321 

t 

0.0 

0.15 

0.832 

2.01 

3.55 

5.38 

7.30 

11.0 

e 

-0.064 

0.1953 

0.439 

0.65 

0.815 

0.923 

0.967 

0.86 

P 

0.0 

0  .  0293 

0.365 

1.308 

2.89 

4.97 

7.05 

9.45 

0-01 

150 

180 

210 

240 

270 

300 

330 

360 

0° 

146°  20' 

176°  20' 

206°  20' 

236°  20' 

266°  20' 

296°  20' 

326°  20' 

356°  20' 

sin  0 

0.5544 

0.064 

-0.4436 

-0.8323 

-0.998 

-0.8962 

-0.5544 

-0.064 

a  =  0  —  86°  20' 

60° 

90° 

120° 

150° 

180° 

210° 

240° 

270° 

sin  a 

0.866 

1.0 

0.866 

0.5 

0.0 

-0.5 

-0.866 

-1.0 

0(rad.) 

2.55 

3.08 

3.6 

4.125 

4.65 

5.17 

5.7 

6.22 

0  +  0.064 

2.614 

3.144 

3.664 

4.189 

4.714 

5.234 

5.764 

6.284 

X 

0.05228 

0.06288 

0.07328 

0.08378 

0.09428 

0.10468 

0.11528 

0.12568 

•"« 

0.949 

0.939 

0.929 

0.9195 

0.91 

0.902 

0.891 

0.882 

y 

0.2175 

0.262 

0.305 

0.349 

0.393 

0.436 

0.48 

0.524 

b  -e~" 

0.804 

0.768 

0.737 

0.706 

0.674 

0.65 

0.62 

0.592 

a 

0.822 

0.939 

0.805 

0.46 

0.0 

-0.451 

-0.772 

-0.882 

a  +  b 

1.626 

1.707 

1.542 

1.166 

0.674 

0.199 

0.152 

0.29 

i 

13.53 

14.2 

12.85 

9.7 

5.6 

1.66 

-1.266 

-2.415 

e 

0.525 

0.060 

-0.412 

-0.765 

-0.908 

-0.809 

-0.494 

-0.0565 

P 

7.1 

0.0851 

-5.3 

-7.42 

-5.08 

-1.343 

0.625 

1.364 

Class  A.—  0!  =  41°  20'  =0.718  radian 

i  =  8.32[e-0-02^-°-718>  sin(0  -  86°  20')  -  €-0-0833(0-0.7l8)  sin(_45°)] 
=  8.32[€~z  sin(0  -  86°  20')  +  0.707  e~v] 
e  =  e~x  sin  0 


0-01 

0 

30 

60 

90 

120 

150 

180 

210 

0 

41°  20' 

71°  20' 

101°  20' 

131°  20' 

161°  20' 

191°  20' 

221°  20' 

251°  20' 

•"* 

1 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

sin  0 

0.6604 

0.9474 

0.9805 

0.7509 

0.3201 

-0.1965 

-0.6604 

-0.9474 

a" 

-45° 

-15° 

15° 

45° 

75° 

105° 

135° 

165° 

sin  a" 

-0.707 

-0.2588 

0.2588 

0.707 

0.9659 

0.9659 

0.707 

0.2588 

c~x  sin  a" 

-0.707 

-0.256 

0.2535 

0.685 

0.9255 

0.916 

0.664 

0.2405 

707«-w 

0.707 

0.677 

0.648 

0.621 

0.595 

0.568 

0.543 

0.521 

i 

0.0 

3.5 

7.5 

10.87 

12.67 

12.35 

10.05 

6.34 

e 

0.6604 

0.936 

0.96 

0.728 

0.307 

-0.1864 

-0.62 

-0.88 

P 

0.0 

3.28 

7.2 

7.9 

3.89 

-2.3 

-6.23 

-5.58 

308 


ELECTRICAL  ENGINEERING 


10 


FIG.  229. 


FIG.  230. 


SHORT-CIRCUIT  OF  ALTERNATORS 


309 


10 


FIG.  231. 


Class  A.—0!  =  86°  20'  =  1.50  radians. 
The  equations  (423)  and  (421)  become: 

i  =  8.32€-°-02^  -  J-5)  sin  (6  -  86°  20')  =  8.32e-x  sin  a, 
e  =  €-0-°2(0  -  1.6)  sin  e  =  €-x  sin  Q 

Tabulating : 


9  -  0i 

0 

30 

60 

90 

120 

150 

180 

210 

0° 

86°  20' 

116°  20' 

146°  20' 

176°  20' 

206°  20' 

236°  20' 

266°  20' 

296°  20' 

sin  0 

0.998 

0  .  8962 

0.5544 

0.064 

-0.4436 

-0.8323 

-0.998 

-0.8962 

a° 

0.0 

30.0 

60.0 

90.0 

120.0 

150.0 

180.0 

210.0 

sin  a 

0.0 

0.5 

0.866 

1.0 

0.866 

0.5 

0.0 

-0.5 

0(rad.) 

1.5 

2.03 

2.55 

3.07 

3.60 

4.12 

4.64 

5.17 

0  -  1.5 

0.0 

0.53 

1.05 

1.57 

2.10 

2.62 

3.14 

3.67 

X 

0.0 

0.0106 

0.021 

0.0314 

0.042 

0.0524 

0.0628 

0.0734 

e~x 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

e~x  sin  a 

0.0 

0.4945 

0.848 

0.969 

0.831 

0.4745 

0.0 

-0.4645 

I 

0.0 

4.11 

7.05 

8.06 

6.92 

3.94 

0.0 

-3.86 

e 

0.998 

0.886 

0.542 

0.062 

-0.425 

-0.79 

-0.937 

-0.833 

P 

0.0 

3.64 

3.82 

0.50 

-2.94 

-3.12 

0.0 

3.22 

e  -  0i 

240 

270 

300 

330 

360 

390 

420 

450 

0° 

326°  20' 

356°  20' 

386°  20' 

416°  20' 

446°  20' 

476°  20' 

506°  20' 

536°  20' 

sin  0 

-0.5544 

-0.064 

0.4436 

0.8323 

0.998 

0.8962 

0.5544 

0.064 

a 

240.0 

270.0 

300.0 

330.0 

360.0 

390.0 

420.0 

450.0 

sin  a 

-0.866 

-1.0 

-0.866 

-0.5 

0.0 

0.5 

0.866 

1.0 

0(rad.) 

5.69 

6.21 

6.74 

7.26 

7.78 

8.30 

8.83 

9.35 

0  -  1.5 

4.19 

4.71 

5.24 

5.76 

6.28 

6.80 

7.33 

7.85 

X 

0  .  0838 

0.0942 

0.1048 

0.1152 

0.1256 

0.136 

0.1466 

0.157 

e~x 

0.9194 

0.91 

0.902 

0.89 

0.881 

0.872 

0.864 

0.855 

t~x  sin  a 

-0.796 

-0.91 

-0.782 

-0.445 

0.0 

0.436 

0.749 

0.855 

t 

-6.63 

-7.57 

-6.51 

-3.74 

0.0 

3.63 

6.24 

7.11 

e 

-0.51 

-0.0582 

0.40 

^0.741 

0.880 

0.782 

0.479 

0.0547 

P 

3.38 

0.441 

-2.61 

-2.77 

0.0 

2.84 

2.99 

0.39 

ELECTRICAL  ENGINEERING 


FIG.  232. 

Class  B.--el  =  -  25°  40'  =  -  0.445  radian 

i  =  45.1[e-°-02<'  +  °-445>  sin(0  -  64°  20')  +  6 -0.5(0  +0.445)  j 
=  45.1[e-xsin  (e  -  64°  20')  +  «-*'] 


c-0.02«?  +  0.445)  gi 


sin  6 


e  -  61 


210 


4°  20' 

0.0756 

0. 

-60.0 

-0. 

-0.856 
0.265 
0.767 

-4.015 
0.0748 

-0.30 


34°  20' 
0.564 
0.979 

-30.0 

-0.5 

-0.4895 
0.525 
0.592 
4.625 
0.552 
2.55 


124°  20' 
0.8258 
0.949 

60.0 
0.866 
0.821 
1.31 
0.268 

49.1 
0.784 

38.5 


154°  20' 
0.4331 
0.939 
0 


FIG.  233. 


SHORT-CIRCUIT  OF  ALTERNATORS 


311 


Class  B.—Oi  =  19°  20'  =  0.336  radian 

t  =  45.1[e-°-02<*  -  °-336>  sin  ($  -  64°20')  -  e"0-5^  ~  °-336>  sin(-45°)] 
=  45.1[e-x  sin  (6  -  64°  20')  +  0.707  e~^] 
-  0.336)  =     - 


0-01 

0 

30 

60 

90 

120 

150 

180 

210 

0 

19°  20' 

49°  20' 

79°  20' 

109°  20' 

139°  20' 

169°  20' 

199°  20' 

229°  20' 

sin  0 

0.3311 

0.7585 

0.9827 

0.9436 

0.6517 

0.1851 

-  0.3311 

-0.7587 

«-x 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

t""1  sin  a" 

-0.707 

-0.256 

0.2535 

0.685 

0.9255 

0.916 

0.664 

0.2405 

0.7076-2" 

0.707 

0.542 

0.419 

0.322 

0.2475 

0.1895 

0.147 

0.1118 

t 

0.0 

12.9 

30.35 

45.4 

53.0 

49.9 

36.6 

15.9 

e 

0.3311 

0.75 

0.962 

0.914 

0.625 

0.1758 

-  0.3108 

-0.705 

P 

0.0 

9.67 

29.2 

41.5 

33.1 

8.77 

-11.39 

-11.21 

FIG.  234. 


Class  B.—6i  =  64°  20' 
t  =  45.1e-°-02^  - 


1.12  radian 
2)  sin  (6  -  64°  20') 


=  45.1  e-x,sin  (e  -  64°  20')  =  45.1  <-~x  sin  a' 
e  =  c-o.02(0-i.i2) 


gn 


=    -x 


sn 


e  -  0i 

0 

30 

60 

90 

120 

150 

180 

210 

0 

64°  20' 

94°  20' 

124°  20' 

154°  20' 

184°  20' 

214°  20' 

244°  20' 

274°  20' 

sin  9 

0.9013 

0.9971 

0.8258 

0.4331 

-0.0756 

-0.564 

-0.9013 

-0.9971 

t~x 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

t~x  sin  o' 

0.0 

0.4945 

0.848 

0.969 

0.831 

0.4745 

0.0 

-0.4645 

i 

0.0 

22.3 

38.25 

43.7 

37.5 

21.4 

0.0 

-20.95 

e 

0.9013 

0.985 

0.809 

0.42 

-0.0725 

-0.535 

-0.846 

-0.925 

P 

0.0 

22.0 

30.95 

18.75 

-2.72 

-11.45 

0.0 

+  19.4 

312 


ELECTRICAL  ENGINEERING 


FIG.  235. 

Class  C.—Bi  =  -  8°  20'  =  -  0.145  radian 

i  =  16.5[e-  °-02^  +  °-145>  sin  (0  -  81°  40')  +  6  ~  0-1667(0  +  o.i45)j 

=  16.5[e-  *  sin  (0  -  81°  40')  +  e~  V] 
e  =  e-  0.02(0  +  0.145)  sin  e  =  €-x  sin  0> 


0-01 

0 

30 

60 

90 

120 

150 

180 

210 

0 

-8°  20' 

21°  40' 

51°  40' 

81°  40' 

111°  40' 

141°  40' 

171°  40' 

201°  40' 

sin  0 

-0.1449 

0.3692 

0.7844 

0.9894 

0.9293 

0.6202 

0.1449 

-0.3692 

«-* 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

a" 

-90.0 

-60.0 

-30.0 

0.0 

30.0 

60.0 

90.0 

120.0 

e-*  sin  a" 

-1.0 

-0.856 

-0.49 

0.0 

0.48 

0.821 

0.939 

0.804 

y" 

0.0 

0.0883 

0.175 

0.262 

0.35 

0.437 

0.523 

0.611 

t-v> 

1.0 

0.915 

0.84 

0.769 

0.705 

0.647 

0.593 

0.543 

i 

0.0 

0.973 

5.77 

12.7 

19.55 

24.2 

25.3 

22.2 

e 

-0.1449 

0.365 

0.767 

0.959 

0.89 

0.589 

0.136 

-0.3427 

P 

0.0 

0.355 

4.425 

12.18 

17.4 

14.25 

3.44 

-7.61 

FIG.  236. 


SHORT-CIRCUIT  OF  ALTERNATORS 


313 


Class  C.—6i  =  36°  40'  =  0.637  radian 

t  =  16.5[e-°-02('-°-637>sin  (0  -  81°40')-  6-°-1667^-°-637)Sin  (-  45°)] 


=  16.5[e-*  sin(0  -  81°  40')  +  0.707  c~v"] 
e  =  c-o.02(*-o.637)  sin  e  =  €-x  sin  ^ 


e  -  Oi 

0 

30 

60 

90 

120 

150 

180 

210 

0 

36°  40' 

66°  40' 

96°  40' 

126°  40' 

156°  40' 

186°  40' 

216°  40' 

246°  40' 

sin  6 

0.5972 

0.9182 

0.9932 

0.8021 

0.3961 

-0.1161 

-0.5972 

-0.9182 

r9 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

e~x  sin  a" 

-0.707 

-0.256 

0.254 

0.685 

0.926 

0.916 

0.664 

0.241 

Q.7Q7<TV" 

0.707 

0.647 

0.594 

0.544 

0.498 

0.457 

0.42 

0.384 

i 

0.0 

6.45 

14.0 

20.25 

23.5 

22.65 

17.9 

10.3 

e 

0.5972 

0.908 

0.972 

0.777 

0.38 

-0.1102 

-0.5605 

-0.853 

P 

0.0 

5.85 

13.6 

15.74 

8.93 

-2.5 

-10.05 

-8.78 

FIG.  237. 

Class  C.—d1  =  81°  40'  =  1.42  radians 

i  =  16.5  €-o.02(0-l.42)  sin  (0  _  glo  40,} 

=  16.5  «-*  sin  (6  -  81°  40')  =  16.5  e~x  sin  a' 
e  =  €-o.02(0-i.42)  sin  e  =  e-*sme. 


e  -  Oi 

0 

30 

60 

90 

120 

150 

180 

210 

0 

81°  40' 

111°  40' 

141°  40' 

171°  40' 

201°  40' 

231°  40' 

261°  40' 

291°  40' 

sin  0 

0.9894 

0.9293 

0.6202 

0.1449 

-0.3692 

-0.7844 

-0.9894 

-0.9293 

•-« 

1.0 

0.989 

0.979 

0.969 

0.959 

0.949 

0.939 

0.929 

€~x  sin  «' 

0.0 

0.495 

0.848 

0.969 

0.831 

0.475 

0.0 

-0.465 

i 

0.0 

8.16 

14.0 

16.0 

13.7 

7.84 

0.0 

-7.67 

e 

0.9894 

0.919 

0.607 

0.1403 

-0.354 

-0.744 

-0.928 

-0.862 

P 

0.0 

7.5 

8.5 

2.245 

-4.85 

-5.83 

0.0 

6.61 

It  is  important,  in  connection  with  the  study  of  short-circuits, 
to  determine  how  great  will  be  the  stress  placed  upon  the  shaft  of 
the  alternator.  From  the  present  calculations  of  power  (class  A), 


314  ELECTRICAL  ENGINEERING 

the  maximum  value  obtained  was  found  to  be  9.5  times  normal. 
As  an  example,  let  the  normal  maximum  output  rating  of  the 
machine  be  assumed  as  10,000  kw.  Then  the  maximum  power 
developed  under  short-circuit  would  be  9.5  X  10,000  =  95,000 
kw. 

A  portion  of  this  power  will  be  supplied  from  the  stored  electro- 
magnetic energy  of  the  field,  and  the  remainder  must  come  from 
the  stored  mechanical  energy,  or  from  the  shaft.  Before  short- 
circuiting,  the  stored  electromagnetic  energy  amounts  to  %Li2, 
where  L  is  the  inductance  of  the  field  system  and  i  is  the  field 
current. 

Since 

L  =  i  X  108' 
the  energy  is, 

w  —  f^  TAS"  ioules- 

Since  it  has  been  assumed  that  the  flux  at  any  instant  is  de- 
termined by  the  equation 


the  energy  given  out  during  any  period  of  time  is 


which  may  be  determined  from  the  known  constants. 
As  an  example,  let 

$  =  150  X  106  lines  of  flux  per  pole, 

n  =  300  turns  per  pole, 

i   =  100  amp.  field  current. 
Then 

300  X  150  X  106 
L  =  -     1QO  x  1Q8  —  =  4.5  henrys  per  pole. 

If  all  the  flux  is  destroyed  the  energy  is 

W  =  4  X  lALi*  =  4  X  0.5  X  4.5  X  10,000  =  90,000  joules, 

or  90  kw.  sec. 

If  this  energy  all  disappears  in  Hs  sec->  tne  average  power 
during  this  short  interval  is 

90  X  25  =  2250  kw., 
which  is  furnished  by  the  destruction  of  the  flux. 


SHORT-CIRCUIT  OF  ALTERNATORS  315 

The  total  heat  developed  is  i2Rt,  or 

W  =   CpRdt  =  Ci2RdO, 

where  ti  and  0i  are  used  to  designate  the  initial  moment  of  short- 
circuit,  and  t  and  0  any  subsequent  moment.  If,  for  instance, 
6  —  0i  is  made  equal  to  2?m,  w  is  the  heat  generated  in  n  cycles. 
The  complete  expression  for  the  heat  developed  is  obtained  as 
follows.  From  (119), 

w  =  fi*Rdd  =  R^f  f)  'yV2"'^'0  sin'  (0-0)  _&-«!(«-*) 
sin  (0  -  0)  sin  (0!  -  0)  +  e-2«<«-*>  gin2  (0!  -  p)]dO,  where  «0  is 

written  for  —  >  a  for  -»  and  «i  for  —  +  -• 
#o  a;  XQ       x 

Carrying  out  the  integration,  this  becomes 

sin  2(0  -  0)  -  q0  cos  2(0  -  0) 


,  2<M(9_9l) 

4\  x    Zl 


p? 


o:o2  +  1 
~'         ">- 


The  maximum  heat  is  produced  when  the  short-circuit  occurs 
at  such  a  time  that  sin  (0i  —  /3)  =  —  1.  The  maximum  heat 
produced  in  n  cycles  is  then: 


W  =     PRdO  =  R  —  ^  -  +  - — ^ 

,  \  a?   Z/  L       4a0  2« 

—  ..    , ai  2(1  —  €-2™ai)J  approx.     (120) 
The  average  power  developed  during  n  cycles  is 


E  I 

Since  the  rated  power  of  an  alternator  is     ™  m>  the  ratio 

Power  during  short-circuit  _  2^n  ^  ^     °  W 


=   *rav, 


rated  power  Emlm  Emlmirn 

2 

1?    T  1 

On  the  percentage  basis,     ^  m  =  ^  or  if  E7  =  1,  where  effect- 
ive values  of  voltage  and  current  are  used,  the  ratio  becomes  ^ — 


316 


ELECTRICAL  ENGINEERING 


Problem  97. — Calculate  and  plot  the  ratio  of  average  power,  under  the 
condition  of  maximum  heat  (Eq.  120),  to  rated  power,  for  values  of  n  from 
1  to  10,  for  the  three  classes  of  alternators. 

Class  A. — From  the  previous  calculation  (page  315), 


f)  2  =  °-01(8-32)2  =  °-692 


0.692       0.11 


a0  =  0.02, 
a    =  0.0833, 
ai  =  0.1033, 

«!2  =  0.0107, 


4a0  =  0.08, 
2a    =  0.1667, 
2«!  =  0.2067, 

-  ax2  =  1.0107, 


47ra0  =  0.2515 
4™    =  1.048 
27rai  =0.65 

2-  -  0.2043. 


Supplying  these  values  (121)  becomes 

_  £-1.408/» 


0.11  /I  -  €-0-2515n 

=      n    \        "  ^° 


0.08 


0.1667 


-  0.2043(1  - 


2.01254 


n 


1*375   -0.2515n 

c 

n 


0.66  _ 

€ 


0.02246  _0.65n 
n      € 


=  a  —  b  —  c  -}-  d. 
Tabulating : 


n 

1 

2 

3 

4 

5 

6 

8 

10 

2.01254 

2ni  occ/i 

IrtAAOT 

OATAQO 

OKAO1  A 

OA  HOK1 

000  SA  O 

OOK1  K'T 

0=3        n 

.UI^5O4 

.UVoZf 

.  o7Uoo 

.  oUo!4 

.4UZ51 

.  ooo4.<2 

.  ZOluf 

0.  20125 

1.375 

1.375 

0.6875 

0.4583 

0.3438 

0.275 

0.2297 

0.1719 

0.1375 

n 

0.66 

n 

0.66 

0.33 

0.22 

0.165 

0.132 

0.11 

0.0825 

0.066 

0.02246 

Onoo/m 

f\     /\i  i  OO 

OAAT/lfi 

OAAKftO 

Of\f\A  >ino 

Of\f\O*7A 

OAAOO1 

OAAOO>t  A 

n 

.  UZZ4O 

U.Ull^o 

.UO749 

.  UUOD  £i 

.  UU44y  2 

.UUo74 

.UU2ol 

.  002240 

0.2515n 

0.2515 

0.503 

0.7545 

1.006 

1.2575 

1.509 

2.012 

2.515 

e-o.2s,8n 

0.778 

0.605 

0.47 

0.364 

0.284 

0.22 

0.134 

0.081 

1.048n 

1.048 

2.096 

3.144 

4.192 

5.24 

6.288 

8.384 

10.48 

€-1.048n 

0.35 

0.124 

0.043 

0.015 

0.0058 

0.0016 

0.0 

0.0 

0.65n 

0.65 

1.3 

1.95 

2.6 

3.25 

3.9 

5.2 

6.5 

€-0..5n 

0.523 

0.272 

0.142 

0.074 

0.038 

0.019 

0.006 

0.0012 

b 

1.07 

0.416 

0.227 

0.125 

0.078 

0.0505 

0.023 

0.01115 

C 

0.231 

0.041 

0.00945 

0  .  00247 

0.000765 

0.000176 

0.0 

0.0 

d 

0.01175 

0.003055 

0.001063 

0.000416 

0.0001707 

0.000071 

0.000017 

0.0000027 

Pa,. 

0.7233 

0.5523 

0.4354 

0.3761 

0.3239 

0.2848 

0  .  2286 

0.1901 

Ratio 

1.4466 

1  .  1046 

0  .  8708 

0.7522 

0.6478 

0  .  5696 

0.4572 

0  .  3802 

Class  B.— 


a0  =  0.02, 
a    =  0.5, 
«i  =  0.52, 

a!2  =  0.271, 


=  0.01  (45.1)2  =  20.34 
20.34  =  3.24 

2irn          n 
4«0  =  0.08, 
2*    =  1.0, 
2«i  =  1.04, 

«i2  =  1.271, 


47ra0  =  0.2515. 
4ira    =  6.28. 
=  3.267. 

=  0.817. 


SHORT-CIRCUIT  OF  ALTERNATORS 


317 


Supplying  values  (121)  becomes 

'm'  =  "^TL        (U)8~~  ~T 

41.09       40.5  _0.2515n      3.24     *28n  .   2.65  _3.267n 


-  0.817(1  - 


n 


n 


=  a'  _  5'  -  c' 
Tabulating: 


n 

1 

2 

3 

4 

5 

6 

8 

10 

a' 

41.09 

20.54 

13.7 

10.27 

8.22 

6.85 

5.135 

4.109 

40.5 

n 

40.5 

20.25 

13.5 

10.125 

8.1 

6.75 

5.063 

4.05 

3.24 

3.24 

1.62 

1.08 

0.81 

0.648 

0.54 

0.405 

0.324 

n 

2.65 

2.65 

1.325 

0.883 

0.663 

0.53 

0.442 

0.331 

0.265 

n 

c—  0.2516n 

0.778 

0.605 

0.47 

0.364 

0.284 

0.22 

0.134 

0.081 

c-6.28n 

0.0016 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

g—  3.267n 

0.038 

0.0017 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

6' 

31.53 

12.25 

6.34 

^.688 

2.3 

1.485 

0.678 

0.328 

c' 

0.00518 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

d' 

0.101 

0.00225 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

Pa,. 

9.66 

8.29 

7.36 

6.58 

5.92 

5.37 

4.46 

3.78 

Ratio 

19.32 

16.58 

14.72 

13.16 

11.84 

10.74 

8.92 

7.56 

Class  C.— 


(jjf     y\  2 
-~  j)     -  0.01(16.5)2  =  2.7225 

2.7225      0.434 


«0  =  0.02, 
«    =  0.1667, 
ai  =  0.1867, 

«i2  =  0.035, 


27m  n 

4a0  =0.08, 
2a    =  0.333, 
2ai  =  0.3733, 

1  +  «i2  =  1.035, 


47ra0  =  0.2515. 
4™  =  2.093. 
27rai  =  1.173. 

2ai       =0.361. 


1  + 


Supplying  values  (121)  becomes 
0.434  r 

•L    av  *    ~^ 

n 


0.434  rl  -  e-o.25i6» 


0.08 

6.565      5.42  _0.2515n 
c 

n  n 

=  a"  -  V  -  c"  +  d". 


-1.173n 


>i 


318  ELECTRICAL  ENGINEERING 

Tabulating : 


n 

1 

2 

3 

4 

5 

6 

8 

10 

6.565 

6Cf»  C 

q    ooq 

0     I  CK 

11 

i    qi  q 

i    078 

0090^ 

OfiCCK 

n 

.000 

o  .  Zoo 

—  .  1OO 

l  .  olo 

1  .  \Jl  O 

,  o^uo 

.  OODO 

5.42 

5.42 

2.71 

1.807 

1.355 

1.084 

0.903 

0.677 

0.542 

n 

1.302 

0«K1 

0 

0  326 

OOfiO 

0  217 

n   iAq 

0   1302 

n 

l.oUJ 

.  OOJL 

.  ^UU 

W  .  £  \.  9 

\J  .  XUO 

U  .  JLOU« 

0.157 

0.157 

0.078 

0.052 

0.039 

0.031 

0.026 

0.0195 

0.0157 

n 

^-0.25157* 

0.778 

0.605 

0.47 

0.364 

0.284 

0.22 

0.134 

0.081 

e-2.093n 

0.124 

0.015 

0.0018 

0.0 

0.0 

0.0 

0.0 

0.0 

e-1.173n 

0.308 

0.097 

0.029 

0.0095 

0.0025 

0.0005 

0.0 

0.0 

V 

4.22 

1.64 

0.849 

0.493 

0.308 

0.1988 

0.0906 

0.0439 

c" 

0.1615 

0.00975 

0.00078 

0.0 

0.0 

0.0 

0.0 

0.0 

d" 

0.0483 

0.00756 

0.00151 

0.0 

0.0 

0.0 

0.0 

0.0 

Po,. 

2.23 

1.64 

1.31 

1.15 

1.00 

0.88 

0.73 

0.61 

Ratio 

4.46 

3.28 

2.62 

2.30 

2.00 

1.76 

1.46 

1.22 

20 


18 


16 


14 


12 


10 


xo. 


As  an  illustration  of  the 
power  developed  under  short- 
circuit,  consider  the  genera- 
tor of  class  A.  The  average 
power  developed  in  the  first 
cycle  under  the  worst  condi- 
tion is  found  to  be  0.7233, 
where  Em  =  1,  Im  =  1,  and 
Pav.  =  normal  power  output 


123 


45 
Cycles 


678 


10 


FIG.  238. 


If,  now,  a  25-cycle  machine 
of  5000-kw.  rating  of  this 
type,  is  considered,  the  aver- 
age power  during  the  first 
cycle  is  10,000  X  0.7233  = 
7,233  kw.  The  instantaneous 
maximum  of  power  has 
already  been  found  to  be 
95,000  kw. 

Problem  98.  —  Determine  the 
above  relation  for  the  machines 
of  classes  B  and  C. 


SHORT-CIRCUIT  OF  ALTERNATORS 


319 


Stresses  on  End -connections  of  the  Armature  Coils. — When 
end-connections  run  parallel  for  some  distance,  the  forces  exerted 
on  them  are  often  very  great  at  the  instants  of  heavy  current 
during  short-circuit.  The  force  at  any  time  may  be  determined 
to  a  sufficient  degree  of  approximation  by  multiplying  the  average 
density  of  the  flux  through  one  conductor  due  to  the  other 
conductor,  by  the  current  in  that  conductor. 

Consider  two  similar  conductors  of  radius,  r,  with  a  distance,  d, 
between  centers.  To  find  the  average  flux  through  conductor 
B  due  to  the  current  in  conductor  A .  The  flux  through  any  ele- 
ment, dx,  of  B  is,  per  centimeter  length  of  conductor, 

2Idx 


nut    —        ~  —  — j 

2-n-X  X 

where  I  is  in  abamperes,  and  /*  is  taken  as  unity.     The  average 
flux  density  is  then: 

h'      efo       /,      d  +  r 


—  r 


In  general,  the  force  exerted  is  BIl  dynes,  where  I  is  length  of 
the  wire  in  centimeters 


VI 


.76" 

1 


FIG.  239. 


FIG.  240. 


MM,        t  F       /2  1       d  +  r  j 

1  nus,  force  per  cm.  =  -y  =  —  log  -j  -  -  dynes. 

t  T  d  —  T 

«...   ,  F        P    ,      d+r 

[f  7  is  in  amperes,  y  ==  log 


If  dimensions  are  given  in  inches,  the  formula  remains  the  same. 

Example.  —  Consider  two  adjacent  conductors,  as  shown  in  Fig. 
240.  The  area  of  each  conductor  is  0.2345  sq.  in.  The  current 
density  is  taken  as  2000  amp.  per  sq.  in.  under  normal  conditions. 
Therefore  maximum  normal  current  is 

Im  =  \/2  X  0.2345  X  2000  =  664  amp, 
d  =  0.5,  r  =  0.1562,  I  =  20. 


320   '  ELECTRICAL  ENGINEERING 

The  maximum  force  under  normal  load  is  then 

(664)  2  X  20         0.6562  366,000 

F  =  100  X  0.1562  log  03438  =  366>°00  dynes  =  44^000  = 

0.822  Ib. 

This  shows  that  the  force  under  normal  conditions  is  very 
slight.  Under  short-circuit  the  ratios  of  maximum  current  to 
normal  current  for  the  three  classes  of  machines  considered,  were, 
respectively: 

for  class  A,    14.2 

for  class  B,    52.0 

for  class  C,    25.3. 

Thus,  the  maximum  short-circuit  forces  are,  for  the  three  classes 
under  the  dimensions  assumed: 
FA  (max.)  =  0.822  X  1O2  =    166  Ib. 
FB  (max.)  =  0.822  X  52^_    =  2220  Ib. 
Fc  (max.)  =  0.822  X  25.32  =    527  Ib. 

Problem  99.  —  Discuss  the  effects  of  changing  the  values  of  r  and  d  on 
the  forces  exerted  on  the  end-connections. 

In  general,  the  effect  of  short-circuit  as  obtained  in  machines  of  class  B, 
was  much  decreased  by  the  addition  of  external  reactance,  as  exemplified  in 
class  C.  What  change  in  the  relative  positions  of  the  end-connections  would 
be  necessary  to  reduce  the  force  as  obtained  for  class  B  to  that  of  class  C 
machines? 

Multiphase  Short-circuits.  —  The  voltage  of  any  phase,  m,  of  a 
multiphase,  alternator  in  the  steady  period  of  operation  is  ex- 
pressed by 

_         .        /  .     27TW 

em  =  Em  sin 


Thus,  for  a  three-phase  generator,  the  voltages  are 

e\  =  Em  sin  (ut  +  0) 

62  =  Em  sin  (ut  +  120) 

63  =  Em  sin  (co£  +  240) 

where  m  has  the  values,  0,  1,  and  2,  respectively.  (For  two-phase 
alternators,  n  must  be  taken  as  4,  not  as  2,  since  the  voltages 
differ  by  90°,  not  by  180°.) 

The  currents  of  a  three-phase  alternator  are: 

ii  =  Im  sin  (coZ  +  0) 

it  =  Im  sin  M  +  120  +  0) 

it  =  Im  sin  (co*  +  240  +  0). 


SHORT-CIRCUIT  OF  ALTERNATORS  321 

The  transient  voltage,  for  example,  of  the  second  phase,  is, 
from  (117)  in  which  Em  is  substituted  for  Eim,  and  E2  is  neglected, 
as  in  the  later  calculations, 

62  =  Emsin(6  +  120)€~^°('     9l\ 

Equating  this  to  izr  +  x  ~i  as  previously  done  for  the  single- 

phase  machine,  the  current  during  the  transient  period  is  found 
to  be 


'    sin  (0  +  120  -  ft  -  ' 

sin  (0i  +  120  -ft]- 

A  still  shorter  but  less  close  approximation  is  made  by  con- 
X 

sidering     -  =  1,  and  0  =  90°.    The  current  is  then 


12  =  cos 


120)  -  f-%(e-9*  cos  (B  +  120)]. 


In  a  polyphase  generator,  the  current  for  any  phase  is  given 
approximately  by 


Emr  --(o-ei)        (n    .  -.-  / 

im  =  —  [e  *        ;  cos^i+-^-    -e   *•          cos+^-      (122) 

where  n  is  the  number  of  phases  and  m  has  the  values  0,  1,   2, 

.    (n  -  1). 

Power  developed  in  any  phase,  at  any  instant,  is  the  product  ei. 
The  whole  power  of  a  three-phase  generator  is,  at  any  instant,  the 
sum  of  the  three  products,  e&i,  e2iz,  e3i3,  of  the  individual  phases. 

Problem  100.  —  Perform  the  operation  just  indicated  and  prove  that  the 
power  of  a  three-phase  generator  is 


sin-  (•- 


This  equation  shows  that  power  of  a  polyphase  generator  is 
entirely  independent  of  the  time  of  closing  of  the  switch.  This 
time  may  have  any  value  assigned  to  81,  but  the  time  at  any  in- 
stant after  the  switch  is  closed  is  represented  by  B  —  0i,  which  is 
independent  of  0i. 

This  is  quite  different  from  the  case  of  single-phase  short-cir- 
cuits in  which  the  power,  similarly  determined,  is 
21 


322  ELECTRICAL  ENGINEERING 

E 
P  (one-phase)  = 

cos  Oi  sin  d-Q.5e~2x°<>(0~ei}  sin  20]  • 

In  this  equation,  0  enters  independently  of  0lt  Cos  0i  is,  of 
course,  a  constant. 

From  the  power  equations,  the  torque  on  the  shaft  at  any 
instant  may  be  determined. 

Problem  101.  —  Show  that  the  maximum  power  of  a  single-phase  short- 
circuit  on  a  three-phase  machine  is  two-thirds  of  that  of  a  three-phase 
short-circuit  on  the  same  machine  and  explain  in  words  the  basis  for  this 
relationship. 

Armature  Reaction.  —  For  a  three-phase  generator  in  the  steady 
state  of  operation,  the  armature  reactions  of  the  three  phases 
taken  separately  have  been  found  to  be: 

FA1  =  i^T  cos  0, 
FA2  =  *V 


FA3  =  i,T  cos  (e  +  y)   =  i3T  cos  (0  +  I)  , 
where  T  is  the  number  of  effective  turns  per  phase  and  0,  0  +  -=-> 

o 

4rr 
0  +  -o~  represent  the  angular  space  positions  of  the  armature  core 

with  respect  to  the  field  core.     Substituting  the  values  of  i  from 
(122)  the  transient  values  of  the  armature  reaction  are: 

ErnT[   -'-(6-9!)  -^(0-0i)  /I   +  COS  20\  "| 

--  '  --  2  --  )  \ 


--- 

Al  =    --«   x  cos  0!  cos  0  -  € 


m  -       g-l  2w\  /  27T\ 

FA2  =    —  —   e    sv  cos      i  +          cos     6  +  -  € 


cos 


SHORT-CIRCUIT  OF  ALTERNATORS  323 

Adding  these  three  equations,  the  total  three-phase  armature 
reaction  is: 


cos  fl  _  9i  _ 


Problem  102. — Prove  that  the  armature  reaction  of  a  polyphase  generator 
is: 

ET       rjl  r~  T  7*0  ~~\ 

n   Jl/m-i   I (6  —  0j)  .  .  —  - — •((? — 81)   I          /'1OQ'\ 

2      x     L 

Problem  103. — Plot  single-phase  and  three-phase  armature  reaction 
curves  for  the  alternator  for  which  waves  of  e,  i,  and  p  have  been  derived, 
and  discuss  their  characteristic  differences. 

Electromotive  Force  and  Current  Induced  in  the  Field  Windings. 

— Excessive  voltage  may  be  induced  in  the  field  windings  and 
cause  breakdown  of  insulation.  In  general,  the  induced  voltage 

is  proportional  to  —^7—     It  is,  however,  difficult  to  obtain  a  reliable 

value  of  the  voltage  owing  to  the  fact  that  the  flux  cannot  pene- 
trate uniformly  into  the  magnet  cores  during  the  exceedingly 
short  time  allowed  by  the  rapidly  changing  current. 

The  induced  field  current  may  also  be  abnormally  great.  By 
installing  a  circuit  breaker  in  the  exciter  circuit,  the  rush  of  cur- 
rent may  cause  the  circuit  to  be  opened,  thus  taking  off  the  field 
current  from  the  short-circuited  alternator. 

Example. — Let  the  normal  field  excitation  be  18,000  amp.- 
turns  per  pole,  and  the  normal  armature  reaction  be  9000  amp.- 
turns.  If  the  armature  reactance  is  10  per  cent.,  the  maximum 
short-circuit  current  would  be  approximately  seventeen  times 
normal  current.  The  armature  short-circuit  amp.-turns  are  then 
153,000.  Assuming  20  per  cent,  leakage  between  armature  and 
field,  the  effective  armature  reaction  is 

0.8  X  153,000  =  122,000  amp.-turns  on  the  field  core. 
The  field  current  may  then  attain  the  value  of 

1 22  000 

'        X  normal  =  6.8  X  normal  current. 

If  the  circuit  breaker  is  set  for  twice  normal  current,  it  will 
open  the  circuit. 


CHAPTER  XLI 

SYNCHRONOUS  MOTORS 

When  the  ordinary  alternator  is  supplied  with  electrical  energy 
and  made  to  do  mechanical  work,  it  becomes  a  synchronous 
motor.  The  name  is  meant  to  indicate  its  chief  characteristic, 
namely  that  of  running  in  exact  synchronism  with  the  generator 
which  supplies  it  with  energy.  If  the  frequency  of  the  generator 
is  60  cycles  per  second,  that  of  the  motor — its  counter  e.m.f. — is 
also  60  cycles.  This  condition  is  the  result  of  the  electromag- 
netic relationship  between  the  field  and  armature  cores;  the  field 
core  changes  its  position  in  space  by  means  of  mechanical  rota- 
tion, the  position  of  the  magnetic  field  due  to  the  armature 
magnetomotive  force  changes  in  space  because  of  the  time-phase 
relationships  and  alternation  of  the  currents.  The  driving  force 
of  the  motor  is  maintained  only  by  the  existence  of  a  constant 
relationship  between  the  field  and  armature  m.m.f.  The  rate 
of  rotation  of  the  armature  m.m.f.  is  fixed  by  the  frequency  of 
supply.  The  field  has  no  fixed  rate  of  rotation  of  its  own  and  is 
therefore  free  to  accept  that  imposed  by  the  armature. 

The  operation  of  the  synchronous  motor  may  be  affected  either 
by  changing  its  load  or  by  altering  its  field  excitation.  These 
may  be  called  primary  means  of  adjustment  since  they  are  ap- 
plicable to  any  motor  in  operation.  Since,  however,  the  speed 
cannot  be  changed,  it  becomes  a  matter  of  great  interest  and  also 
of  importance  to  find  out  what  is  changed,  and  what  peculiar 
and  valuable  characteristics  are  associated  with  this  hitherto  un- 
encountered  characteristic  of  synchronous  speed. 

There  are  also  secondary  means  of  adjustment  by  which  varia- 
tion in  the  motor  performance  may  be  brought  about.  These 
involve  changes  in  the  constants  of  the  line  or  the  motor  circuit. 
Thus,  in  the  matter  of  design,  it  is  important  to  study  the  effects 
of  different  values  of  resistance  and  reactance  of  the  armature. 
In  operation,  with  a  constant  generator  terminal  e.m.f.,  resistance 
and  reactance  may  be  inserted  or  withdrawn  from  the  line,  thus 
altering  the  total  r  and  x  of  the  circuit. 

324 


SYNCHRONOUS  MOTORS 


325 


A  thorough  understanding  of  the  effect  of  these  constants  is 
essential  from  a  practical  as  well  as  a  theoretical  point  of  view. 
A  motor  which,  for  instance,  operates  perfectly  satisfactorily  on 
one  line  may  be  entirely  unstable  and  even  unable  to  carry  its 
load  or  even  a  small  fraction  thereof  on  another  line. 

It  will,  for  instance,  be  evident  that 
a  high  resistance  line  tends  to  make 
the  motor  unstable  unless  the  reac- 
tance is  also  considerable.  In  synchro- 
nous motor  operation  a  fair  amount  of 
line  reactance  is  essential;  in  fact,  the 
very  ability  of  the  motor  to  carry  load 
depends  upon  the  presence  of  reactance 
in  the  motor  circuit. 

Let  E  be  the  e.m.f.  counter  generated 
in  the  motor.  The  resultant  flux  will 

then 'be  90°  ahead  of  E.  Assuming  a  current  I,  as  shown  in 
Fig.  241,  this  current  produces  a  m.m.f.  in  phase  with  itself 
and  which  may  be  taken  equal  to  it,  by  choosing  a  suitable 
scale.  The  armature  m.m.f.  thus  produced,  when 
added  vectorially  to  the  field  m.m.f.,  will  produce 
the  resultant  m.m.f.,  which  gives  the  resultant  flux 
<j>r.  <£/  represents  the  direction  of  the  field  m.m.f. 
In  order  to  force  the  current  through  the  im- 
pedance of  the  armature,  it  is  necessary  to  have  an 
e.m.f.  equal  to  the  impedance  drop  IZ.  As  shown 
in  the  figure,  IZ  is  the  voltage  which  overcomes 
the  resistance  and  reactance  of  the  armature.  The  impressed 
voltage,  EQ  must  be  the  vector  sum  of  this  IZ  drop  and  the  voltage 
—  E,  necessary  to  overcome  the  counter  e.m.f.  of  the  motor. 

.   .    EQ    =    Lfj    —    E. 

The  space  relations  indicated  by  Fig. 
241  are  illustrated  by  the  sketch  of  a 
two-pole    machine    in    Fig.    242.     The 
vector    relationship   may   also   be   con- 
sidered from  a  somewhat  different  point  of  view,  illustrated  in 
Fig.  243.     Here  there  are  two  e.m.fs.,  E  and  EQ,  acting  in  a 
circuit  of  impedance  Z. 

If  EQ  is  the  generator  terminal  e.m.f.,  then  z  is  the  combined 


326  ELECTRICAL  ENGINEERING 

impedance  of  the  line  and  the  motor.  The  counter  e.m.f.,  Et 
of  the  motor,  is  naturally  in  a  direction  to  more  or  less  oppose  EQ. 
The  vector  sum  of  EQ  and  E  is  Ez  which  is  the  e.m.f.  which 
actually  overcomes  the  impedance  2,  of  the  circuit.  The  current 

0* 

7  lags  behind  Es  by  an  angle  a,  such  that  tan  a  =  — 
...  •  f 

The  motor  output  is  —  E  X  /  cos  p  =  P. 
This  is  seen  to  be  negative  thus  representing  power  supplied 
to  the  machine  or  motor  action. 

The  generator  input  is  EQ  X  /  cos  q  =  P0. 
The  power  lost  in  the  circuit  is  then  P0  —  P  =  /2r. 
If  a  is  a  large  angle,  representing  large  reactance,  the  motor 
is  more  stable  than  if  a  is  small. 

If  a  is  small,  the  projection  of  I  on 
E  may  even  be  positive,  giving  genera- 
tor power  instead  of  motor  power  in 
which  case  the  motor  cannot  carry  me- 
chanical load.  Oftentimes  poorly  act- 
ing synchronous  motors  may  be  greatly 
benefited  by  increasing  the  angle  a  by 
the  insertion  of  self-inductance  in  the 
line.  For  a  given  load  on  the  motor/ 
the  angle  7,  between  the  field  m.m.f. 
FIG.  244.  and  the  resultant  m.m.f.  is  almost  con- 

stant.    OF/  evidently  depends  on  both 

the  amount  and  the  phase  of  the  current.  The  counter  e.m.f., 
E,  on  the  other  hand,  is  fairly  constant  for  all  loads.  It  de- 
pends on  the  actual  resultant  flux  in  the  air  gap  which  is  fairly 
constant  for  all  loads.  For  constant  motor  load,  P  —  oE  X  0/0 
(Fig.  244)  and  the  locus  of  the  ends  of  the  current  vectors  will 
be  along  the  dotted  lines  HQ.  The  corresponding  locus  of  field 
flux  vectors  will  be  along  F/Fo.  If,  however,  the  angle  7  is 
assumed  constant  the  two  locii  will  be  IFr  and  OF/,  for  vary- 
ing field  excitation.  But  this  condition  will  correspond  to  a 
variable  load.  If  OFr  is  great  with  respect  to  01 — that  is,  if 
the  angle  7  is  small — the  variation  of  both  7  and  the  motor 
power  is  small  for  a  considerable  variation  of  the  field  flux 
about  the  normal  value.  Plotting  the  armature  current  against 
the  field  or  the  field  current,  gives  the  familiar  "  F-curves." 


SYNCHRONOUS  MOTORS 


327 


As  E  cannot  be  assumed  constant,  especially  where  r  or  x 
is  large,  the  condition  of  constant  power  output  cannot  be 
shown  by  the  above  vector  diagrams,  since  the  power  is  not 
represented  by  a  constant  projection  of  /  on  the  horizontal. 
Constant  power  input  may,  however,  be  assumed  with  constant 
eQ  impressed,  and  the  power  input  is  then  proportional  to  the 
projection  of  I  on  e0,  Fig.  245.  Moreover,  constant  power 
input,  over  a  considerable  range  of  current  on  both  sides  of  the 
minimum,  is  approximately  constant  power  output,  since  the 
difference  is  only  Pr  which  is  small  and  which  may  have  small 
variation.  It  is  readily  possible,  therefore,  to  calculate  E  for 
constant  power  input,  since 

\(eQ  -  IZ  cos  (0  +  a))2  +  (IZ  sin  (0  +  a))2 


cos  (0  +  a)  +  PZ2 


E 


-  2e0(ri  -  xi'), 


and  this  may  be  determined  for 
varying  /,  since  i  is  constant  and 
known.  Thus  for  any  input, 


P,-  =  e0i,  and  i'  = 


FIG.  245. 


m* 

FIG.  246. 


Synchronous    Motor    Equations.  —  Assuming    e,    the    motor 
counter  e.m.f.  to  be  the  zero  vector, 


E0  =  —  Eo  cos  ]8  —  JEQ  sin  jS. 

By  using  0  the  minus  sign  is  introduced  into  the  equation  since 
the  true  angle  is  180°  +  /3. 

.*.  Iz  =  e  —  Eo  cos  0  —  jE0  sin  0. 

•  .  ••  • 

and 


(124) 


r  +  jx 


328  ELECTRICAL  ENGINEERING 

Also, 

/  =  i  +  ji' 
whence, 

.       e  —  EQ  cos  g 


and 


-  (e  cos  a  —  EQ  cos  (a  —  /?)) 


i'  =  -(-£/o  sin  (a  —  ft  —  e  sin  cr) 


(125) 


These  values  are  obtained  by  clearing  the  denominator  of  (124) 

T  X 

of  imaginaries  and  remembering  that  cos  a  =  -  and  sin  a.  =  -. 

Mechanical,  or  motor  power  P    =  —  ie. 
Hence  the  generated  power   P0  =  —  ie  +  I2r. 
Substituting  the  values  in  (125)  above,  mechanical  power  = 

P  =  ~z(EQ  cos  (a  -  ft  -  e  cos  a)  (126) 

If  /?  =  0  and  $0  =  e,  P  =  0  or  there  is  no  mechanical  power. 
Also  when  0  =  2a,  P  =  0. 

To  determine  the  maximum  output  (126)  may  be  differentiated 
with  respect  to  0  and  the  result  equated  to  zero.  Thus, 

dP  e 

—  =  0  =  -#0sin(a  -  ft. 

In  this,  sin  (a  —  ft  must  equal  zero,  since  -  EQ  is  not  zero. 

This  gives  a  =  0. 

Hence,  the  power  is  maximum  f or  <*  =  j8  and  is  zero  f  or  /3  =  0 
and  0  =  2a. 

If  0  is  negative,  there  is  generator  action,  or  the  motor  acts  as 
generator. 

When  EQ  and  e  are  unequal,  the  limits  of  /3  are  somewhat 
altered. 

Problem  104. — Given: 

Section  A,  E0  =  1.1,  e  =  1 
Section  B,  E0  =  1,  e  =  1 
Section  C,  E0  =  0.9,  e  =  1 

Assume  the  generator  bus  bars  kept  at  constant  voltage — not  constant 
generator  field  excitation.  The  synchronous  motor  armature  has  2  per 
cent,  resistance,  10  per  cent,  reactance. 


SYNCHRONOUS  MOTORS 


329 


1.  An  overhead  line  connecting  the  machine  has  8  per  cent,  resistance  and 
20  per  cent,  reactance,  all  referred  to  motor.    Constants  will  then  be,  r  =  0.1, 
x  =  0.3,  tan  a.  =  3,  rated  power  =  1.0  =  P. 

2.  An  underground  cable  connecting  the  machine  has  a  high  resistance 
of  18  per  cent,  and  has  negligible  reactance.     The  constants  will  then  be 
r  =  0.2,  x  =  0.1.     Find  for  the  two  cases,  power  output,  total  current,  and 
power  factor  of  the  generator,  and  plot  against  /3  (Fig.  246). 

3.  Find  the  maximum  output,  for  variable  r,  with  (a)  x  =  0.1  (6)  x  =  0.2 
and  plot. 

[(From  Eq.  126, 

Pmax.  =~(Eo   —  €  COS  a)] 

Solution  of  the  first  case.     Section  A. 

#„  =  1.1;  E  =  1;  r  =  0.02  +  0.08  =  0.1; 

x  =  0.10  +  0.20  =  0.30;     tan  a  =  ^  =  3;  a  =  72°: 
E 


Mech.  power,  P  —      [E0  cos1  (a  -  0)  -  E  cos  a]  watts 
E 
Z 


Z  =  A/0.32  +  O.I2  =  0.316;  f  =  3.16;  E  cos  a  =  0.309. 


r 

0 

5 

10 

20 

30 

40 

50 

60 

a 

a  -  ft 

72.0 

67.0 

62.0 

52.0 

42.0 

32.0 

22.0 

12.0 

0.0 

Cos  (a  -  0) 

0.309 

0.391 

0.469 

0.616 

0.743 

0.848 

0.927 

0.978 

1.0 

Eo  cos  (a  —  0) 

0.34 

0.43 

0.516 

0.677 

0.818 

0.933 

1.02 

1.075 

1.1 

—  E  cos  a 

0.031 

0.121 

0.207 

0.368 

0.509 

0.624 

0.711 

0.766 

0.791 

P 

0.098 

0.383 

0.655 

1.16 

1.61 

1.97 

2.25 

2.42 

2.5 

Current  =  J  =  — 


-  2EE0  cos  ft 


Cos/3 

1.0 

0.996 

0.985 

0.94 

0.866 

0.766 

0.643 

0.5 

0.309 

2EEo  cos  /3 

2.2 

2.19 

2.165 

2.065 

1.91 

1.685 

1.415 

1.1 

0.68 

Eo2  +E*-  2EEo  cos/3 

0.01 

0.02 

0.045 

0.145 

0.3 

0.52 

0.79 

1.11 

1.53 

#02  +#2_  2EEo  cos/3 

0.1 

0.141 

0.212 

0.381 

0.547 

0.72 

0.89 

1.05 

1.24 

/ 

0.316 

0.445 

0.67 

1.2 

1.73 

2.28 

2.82 

3.32 

3.92 

Power  factor 


P  +  I*r 
Eol 


gen.   power 
Eol 


IV 

0.01 

0.0198 

0.045 

0.144 

0.3 

0.52 

0.795 

1.1 

1.54 

Gen.  power 

0.108 

0.403 

0.7 

1.304 

1.91 

2.49 

3.045 

3.52 

4.04 

Eol 

0.348 

0.49 

0.737 

1.32 

1.91 

2.51 

3.1 

3.66 

4.31 

P.F. 

0.31 

0.822 

0.947 

0.988 

1.0 

0.994 

0.982 

0.963 

0.937 

330 


ELECTRICAL  ENGINEERING 


Section  B.    E0  =  1 

P  =  3.16[cos  (a  -  /3)  -  0.309] 


P 

0 

5 

10 

20 

30 

40 

50 

60 

a° 

Cos  (a  -  0) 

0.309 

0.391 

0.469 

0.616 

0.743 

0.848 

0.927 

0.978 

1.0 

Cos  (a  -  0)   -  0.309 

0.0 

0.082 

0.160 

0.307 

0.434 

0.539 

0.618 

0.669 

0.691 

P 

0.0 

0.259 

0.505 

0.97 

1.37 

1.70 

1/95 

2.11 

2.18 

/  =  3.16  V2  -  2  cos  0 


2  cos  0 

2.0 

1.992 

1.97 

1.88 

1.732 

1.532 

1.286 

1.0 

0.618 

2  —  2  cos  0 

0.0 

0.008 

0.03 

0.12 

0.268 

0.468 

0.714 

1.0 

1.382 

V2  —  2  cos  0 

0.0 

0  .  0893 

0.173 

0.346 

0.518 

0.684 

0.845 

1.0 

1.175 

I 

0.0 

0.282 

0.547 

1.094 

1.638 

2.16 

2.67 

3.16 

3.71 

Power  factor  = 


P  +  0.1/2 


0.1/2 

0.0 

0.008 

0.03 

0.12 

0.269 

0.468 

0.714 

1.0 

1.38 

P  +  0.1/2 

0.0 

0.267 

0.535 

1.09 

1.639 

2.168 

2.664 

3.11 

3.56 

P.F. 

0.947 

0.978 

0.995 

1.0 

1.0 

0.998 

0.984 

0.96 

Section  C.    E0  =  0.9 

P  -  3.16  [0.9  cos  (a  -  0)  -  0.309] 


0° 

0 

5 

10 

20 

30 

40 

50 

60 

a 

0.9  cos  (a  -  0) 

0.278 

0.352 

0.422 

0.555 

0.669 

0.763 

0.835 

0.88 

0.9 

0.9  cos  (a  -0)  -0.309 

-0.031 

0.043 

0.113 

0.246 

0.36 

0.454 

0.526 

0.571 

0.591 

P 

-0.098 

0.136 

0.357 

0.777 

1.138 

1.435 

1.662 

1.805 

1.87 

/  =  3.16  Vl.81  -  1.8  cos  0 


1.8  cos  0 

1.8 

1.792 

1.772 

1.691 

1.56 

1.38 

1.158 

0.9 

0.556 

1.81  —  1.8  cos  0 

0.01 

0.018 

0.038 

0.119 

0.25 

0.43 

0.652 

0.91 

1.254 

Vl.81  -  1.8  cos  0 

0.1 

0.134 

0.195 

0.345 

0.5 

0.655 

0.807 

0.954 

1.12 

/ 

0.316 

0.423 

0.616 

1.09 

1.58 

2.07 

2.55 

3.015 

3.54 

P.F. 


P  +  O.I/* 
0.9/ 


O.I/* 

0.01 

0.0179 

0.038 

0.119 

0.25 

0.43 

0.652 

0.91 

1.254 

P  +  O.I/a 

-0.088 

0.1539 

0.395 

0.896 

1.388 

1.865 

2.314 

2.715 

3.124 

P.F. 

0.309 

0.405 

0.712 

0.913 

0.975 

1.0 

1.0 

1.0 

0.98 

SYNCHRONOUS  MOTORS 


331 


2nd  Case.     Section  A. 


0.1 


E0  =  1.1,  e  =  1,  r  =  0.2,  x  =  0.1,  tan  a  =  ^  =  0.5,  a  =  26°  36' 


Z  =  V0.04  +  0.01  =  0.2236,  -  =  4.47,  e  cos  a  =  0.8942 

P  =  ^  [EQ  cos  (a  -  0)  -  e  cos  a]  =  4.47[1.1  cos  (a  -  0)  -  0.8942]. 


0° 

0 

2.5 

5 

10 

15 

20 

a 

a  -0 

26.6 

24.1 

21.6 

16.6 

11.6 

6.6 

0.0 

Cos  (a  -  0) 

0.8942 

0.9128 

0.9298 

0.9584 

0.9796 

0.9934 

1.0 

Eo  cos  (a  —  0) 

0.984 

1.004 

1.022 

1.055 

1.078 

1.093 

1.1 

[    ] 

0.0898 

0.1099 

0.1278 

0.1608 

0.1838 

0.1988 

0  .  2058 

P 

0.401 

0.490 

0.571 

0.719 

0.822 

0.889 

0.92 

I.I 

z 


+  e2  -  2eEo  cos  ft  -  4.47  \/2.21  -  2.2  cos  /3 


COS0 

1.0 

0.999 

0.9962 

0  .  9848 

0.9659 

0.9397 

0.8942 

2.2  cos  0 

2.2 

2.198 

2.192 

2.165 

2.126 

2.067 

1.967 

2.21  -  2.2  cos  0 

0.01 

0.012 

0.018 

0.045 

0.084 

0.143 

0.243 

V2.21  -  2.2  cos  0 

0.1 

0.1093 

0.134 

0.212 

0.29 

0.378 

0.493 

I 

0.447 

0.489 

0.599 

0.948 

1.297 

1.69 

2.204 

P.F. 


P  +  I*r       P  +  0.2/2 


Eol 


1.11 


0.2/2 

0.04 

0.048 

0.072 

0.18 

0.337 

0.572 

0.975 

P  +  0.2/2 

0.441 

0.538 

0.643 

0.899 

1.159 

1.461 

1.895 

1.11 

0.4915 

0.538 

0.658 

1.043 

1.426 

1.86 

2.424 

P.F. 

0.897 

1.00 

0.978 

0.861 

0.811 

0.785 

0.781 

2nd  Case.     Section  B. 

E0  =  1;  P  =  4.47[cos  («  -  0)  -  0.8942] 


00 

0 

2.5 

5 

10 

15 

20 

a 

Cos  (a  -  0) 

0  .  8942 

0.9128 

0.9298 

0.9584 

0.9796 

0.9934 

1.0 

[    ] 

0.0 

0.0186 

0.0356 

0.0642 

0.0854 

0.0992 

0  .  1058 

P 

0.0 

0.0832 

0.1592 

0.287 

0.382 

0.443 

0.473 

/  =  4.47\/2  -  2  cos  0 


2  cos0 

2.0 

1.998 

1  .  9924 

1.9696 

1.9318 

1.8794 

1  .  7884 

2  —  2  cos  0 

0.0 

0.002 

0.0076 

0.0304 

0.0682 

0.1206 

0.2116 

\/2  —  2  cos  0 

0.0 

0.0447 

0.0871 

0.174 

0.261 

0.347 

0.46 

/ 

0.0 

0.2 

0.39 

0.779 

1.168 

1.552 

2.06 

332 


ELECTRICAL  ENGINEERING 


P.F. 


P  +  0.2/2 


0.27* 

0.0 

0.008 

0.0305 

0.1216 

0.273 

0.483 

0.85 

P  +  0.2/2 

0.0 

0.0912 

0.1897 

0.4086 

0.655 

0.926 

1.323 

P.F. 



0.456 

0.485 

0.525 

0.561 

0.597 

0.643 

1 

2nd  Case.     Section  C. 

E0  =  0.9;  P  =  4.47[0.9  cos  («  -  0)  -  0.8942] 


0° 

0 

2.5 

5 

10 

15 

20 

a 

0.9  cos  (a  -  0) 
[    ] 
P 

0.805 
-0.0892 
-0.4 

0.822 
-0.0722 
-0.323 

0.836 
-0.0582 
-0.26 

0.863 
-0.0312 
-0.1395 

0.882 
-0.0122 
-0.0545 

0.894 
-0.0002 
-0.00894 

0.9 
0.0058 
0.026 

/  =  4.47  A/1 .81  -  1.8  cos  0 


1.8  cos  0 

1.8 

1.798 

1.793 

1.772 

1.74 

1  .  692 

1.61 

1.81  -  1.8  cos  0 

0.01 

0.012 

0.017 

0.038 

0.07 

0.118 

0.20 

Vl-81  -  1.8cos0 

0.1 

0.1093 

0.1303 

0.1947 

0  .  2645 

0.3435 

0.447 

/ 

0.447 

0.489 

0.583 

0.871 

1.183 

1.535 

2.0 

P.F. 


P  +  0.2/2 
0.9/ 


0.2/2 

0.04 

0.048 

0.068 

0.152 

0.28 

0.472 

0.8 

P  +  0  2/2 

-0.36 

-0.275 

-0.192 

0.0125 

0.2255 

0.463 

0.826 

0.9/ 

0.4025 

0.44 

0.525 

0.785 

1.066 

1.382 

1.8 

P.F. 

-0.894 

-0.625 

-0.3655 

0.0159 

0.2115 

0.335 

0.459 

3d  Case.     Section  A.     E0  =  1.1. 


Max.  mech.  power 


(1) 


=  —  [Eo  —  e  cos  a] 
0.1;  r  =  variable 


-  [1.1  -  COS  a] 


T 

0.025 

0.05 

0.1 

0.2 

0.3 

0.4 

0.6 

2 

0.103 

0.1118 

0.141 

0.223 

0.316 

0.412 

0.608 

Cos  a 

0.2425 

0.447 

0.707 

0.897 

0.95 

0.972 

0.987 

1.1  —  cos  a 

0.8575 

0.653 

0.392 

0.203 

0.15 

0.128 

0.113 

Pm 

8.32 

5.85 

2.78 

0.91 

0.475 

0.311 

0.186 

(2)  x  =  0.2 


z 

0.2015 

0.206 

0.223 

0.273 

0.36 

0.447 

0.632 

Cos  a 

0.124 

0.2426 

0.448 

0.733 

0.833 

0.895 

0.95 

1.1  —  cos  a 

0.976 

0  .  8574 

0.652 

0.367 

0.267 

0.205 

0.15 

Pm 

4.85 

4.16 

2.925 

1.345 

0.741 

0.458 

0.2375 

SYNCHRONOUS  MOTORS 


333 


3d  Case.     Section  B.     E0  =  1. 


(1)    x  =  0.1 ;  Pm  =  -  (1  -  cos  a) 


1  —  cos  a 
pm 

0.7575 
7.35 

0.553 
4.95 

0.293 
2.07 

0.103 
0.462 

0.05 
0.158 

0.028 
0.068 

0.013 
0.0214 

(2)  x    =  0.2 


1   —  COS  a 
Pm 

0.876 
4.35 

0.7574 
3.68 

0.552 
2.475 

0.267 
0.978 

0.167 
0.463 

0.105 
0.235 

0.05 
0.079 

3d  Case.     Section  C.    E0  =  0.9 

(1)  x  -  0. 


0.9  —  cos  a 
pm 

0  .  6575 
6.38 

0.453 
4.05 

0.193 
1.365 

0.003 
0.01346 

-0.05 
-0.158 

-0.072 
-0.175 

-0.087 
-0.143 

(2)  x  =  0.2 


0.9  —  cos  a 
Pm 

0.776 
3.85 

0.6574 
3.19 

0.452 
2.015 

0.167 
0.611 

0.067 
0.1862 

0.005 
0.0112 

-0.05 
-0.0791 

Figs.  247,  248  and  249  show  the  curves  plotted  for  the  three  cases  of  the 
problem. 

CO 


40  50  60  70 


334 


ELECTRICAL  ENGINEERING 


It  is  to  be  noted  that  the  angle  /?,  representing  phase  relation- 
ship of  EQ  and  E,  is  used  as  the  independent  variable,  for  con- 
venience only.  The  real  independent  variable  in  the  first  and 
second  cases  of  the  problem  is  power.  It  is  not  so  convenient 
to  choose  power  for  this  calculation  because  its  choice  depends 
upon  the  angle  0;  moreover,  there  would  be  little  interest  in  mak- 
•ing  /3  the  object  of  the  calculation.  The  values  obtained  in  the 


-0.5 


calculations  permit  of  the  plotting  of  other  interesting  curves, 
for  example,  the  performance  curves  of  current  and  power  factor 
against  the  power. 

Problem  106. — From  the  values  obtained,  plot  and  discuss  the  curves  of 
current  and  power  factor  against  the  power  for  the  three  cases  of  constant 
impressed  voltage  and  constant  induced  voltage. 

Problem  106. — Show,  from  a  study  of  the  curves  already  obtained,  that 
when  a  synchronous  motor  refuses  to  operate  satisfactorily  under  load  the 
remedy  for  the  trouble  may  be  found  either  in  decreasing  the  resistance 
(increasing  the  copper  cross-section)  or  in  increasing  the  reactance. 

The  practical  difficulty  of  looking  at  the  motor  from  the  point  of 
view  of  these  problems  is  that  the  induced  voltage,  E,  is  not 


SYNCHRONOUS  MOTORS 


335 


readily  maintained  constant.  E  depends  upon  Fr,  the  resul- 
tant magnetomotive  force,  or,  more  strictly,  upon  the  resultant 
flux  produced  by  Fr,  whereas  only  the  total  field  excitation  Ff  is 
under  external  control. 

The  practical  problem,  therefore,  is  to  vary  Ff  under  the  condi- 
tion of  constant  load,  or  its  converse,  to  vary  the  load  with 


I 


FIG.  249. 


constant  Ff.  In  such  cases  E  will  also  vary,  and  its  value  may  be 
determined  by  calculation.  Curves  plotted  between  the  current 
input,  /,  and  either  E  or  F/,  the  power  of  the  motor  being  con- 
stant, are  called  phase  characteristics.  In  Fig.  250  is  shown  the 
vector  diagram  of  the  synchronous  motor  drawn  to  show  the  cur- 


336 


ELECTRICAL  ENGINEERING 


rent,  /,  as  the  zero  vector.     The  power  factor  is  indicated  by  the 
phase  angle,  5,  between  EQ  and  I.     As  previously  obtained, 

E  =  iz  -  EQ 

=  i(r  +  jx)  —  EQ  cos  8  —  jE0  sin  5 
=  ir  —  EQ  cos  5  +  j(ix  —  EQ  sin  5) 


and 


E  =  V(ir  -  EQ  cos  5)2  +  (ix  -  EQ  sin  5)2 


When  5  is  positive,  the  current  lags. 
To  find  cos  d  and  sin  5:  Remembering 
that  all  values  are  per  phase,  the  me- 
chanical power  is: 

P  =  elec.  power  -  Pr  =  EQI 

cos  6  -  Pr      (127) 


FIG.  250. 


But  P,  EQ  and  r  are  all  known,  being  given  or  assumed. 
Therefore,  solving, 

s       P  +  Pr 
cos  5  = 


and 


EQ! 

sin  5  =  VI  —  cos2  6. 


It  is  then  necessary  merely  to  assume  values  of  7.  Evidently, 
for  constant  power,  7  will  be  minimum  when  cos  6  =  1.  Hence, 
from  (127), 

T7>  \  Tf    2  P 

_  HIQ  /-C/0  f 

Imin.    =    7^;   —   -y^,    "   -' 

Assuming  now  values  of  7,  beginning  with  7m»n.,  cos  5  and  sin 
5,  and  finally  E,  may  be  obtained  and  tabulated  for  each  value 
chosen. 

As  5  may  be  either  plus  or  minus,  both  values  must  be  taken. 

Under  the  conditions  of  test,  E  is  not  known,  but  the  values 
of  the  field  excitation  or  the  field  current  which  is  proportional 
to  it  are  known.  Having  just  obtained  E  by  calculation,  the 
field  excitation,  F/,  is  next  determined  as  was  done  for  the  case 
of  the  generator  (page  225).  In  this  case,  E  terminal  was  the 
zero  vector,  and  F/  was  found  to  be 


-  bC  -  mi)2  +  (aC  - 


SYNCHRONOUS  MOTORS  337 

where 

a  =  e  +  ir  —  i\x 
b    =  ix  +iir 

F 
C   =  -jjf  =  1  (for  convenience) 

rr 

m  ==  —  -  =  0.5  (for  convenience). 

Problem  107.  —  Determine  and   plot  the  phase   characteristics  for  the 
motor  problem  for  the  three  conditions  : 

A.  Eo  =  1.1,  E  =  1 

B.  #o  =  1,  E  =  1 

C.  Eo  =  0.9,  E  =  1 
when 

P  =  1,  r  =  0.1,  x  =  0.3. 

Solution.  —  The  curve  between  e  and  /  is  first  obtained  from  the  equation 
E  =  V(7r  -  Eo  cos  5)2  +  (7z  -  E0  sin  5)2 

by  substituting  values  of  /,  for  which  sin  5  and  cos  5  can  be  determined,  and 
solving  for  E. 

This  curve,  E  vs.  /,  is  plotted. 
i     Next,  the  field  excitation  is  obtained  from  equation 

Ff  =  V(-  bC  -  mi)2  +  (aC  -  mij* 
or,  more  simply, 


Ff  =        (-  mi)*  +  (CE  - 

since  we  deal  directly  with  induced  e.m.f.  as  the  zero  vector,  and  not  with 
the  terminal  voltage  and  IZ  drops.    The  curve  F/  vs.  7  is  then  plotted. 

The  data  given  are: 
Eo  =  (A)  1.1,  (B)  1,  (C)  0.9,  P  =  1,  r  =  0.1,  x  =  0.3,  C  =  1, 

m  =  0.5,  i  =  I  cos  5,  i\  =  I  sin  5. 

To  get  minimum  current,  cos  5  =  1,  sin  5=0. 

(^  r  Ep_      lEp*  _P       1.1       Jl.212         1 

(A)  i  nin.  -  2r     •  \4rz        r  ~  o.2        \  0.04        0.1 

=  5.5  -  V30.3  -  10  =  5.5  -  4.5  =  1. 

(B)  7^.  =  ^2  -  -4^-10  =  5  -  3.87  =  1.13. 


(C)  /„„.  =         -  i  -  10  =  4.5  -  3.24  =  1.3 

Also,  cos  5  =  T  =  —       —  ;  sin  5  =  \/l  -  cos2  5 


(A)  cos  ,  =  jj  =         -  +  0.09097 

(B)cos5  =  j  +0.17 
(C)cosS  =  i^  +0.1117. 


338 


ELECTRICAL  ENGINEERING 


Tabulating,  Case  (A): 


/ 

1.0 

1.1 

1.3 

1.5 

1.8 

2.5 

4.0 

Jr 

0.1 

0.11 

0.13 

0.15 

0.18 

0.25 

0.4 

Ix 

0.3 

0.33 

0.39 

0.45 

0.54 

0.75 

1.2 

0.909 

0.909 

0  826 

0.7 

0.605 

0.505 

0.363 

0.227 

I 

0.909J 

0.0909 

0.1 

0.1181 

0.1363 

0.1637 

0.2272 

0.3633 

Cos  5 

1.0 

0.926 

0.8181 

0.7413 

0  .  6687 

0.5902 

0.5903 

Sin  5 

0.0 

±0.374 

±0.574 

±0.67 

±0.744 

±0.805 

±0.805 

Eo  cos  & 

1.1 

1.02 

0.9 

0.815 

0.735 

0.65 

0.65 

Eo  sin  S 

0.0 

+  0  411 

±0.631 

±0.737 

±0.818 

±0.885 

±0.885 

Ir  —  Eo  cos  5 

-1.0 

-0.91 

-0.77 

-0.665 

-0.555 

-0.4 

-0.25 

(Jr  -  Eo  cos  5)2 

1.0 

0.83 

0.593 

0.442 

0.308 

0.16 

0.0625 

Ix  —  Eo  sin  S 

0.3 

-0.081 

-0.241 

-0.287 

-0.278 

-0.135 

+0.315 

Where   E 

sin  5  is  + 

(Ix  -  Eo  sin  5)2 

0.09 

0.00657 

0.0581 

0.0875 

0.0773 

0.01825 

0.0994 

E* 

1.09 

0.83657 

0.6511 

0.5245 

0.3853 

0.17825 

0.1619 

E 

1.042 

0.914 

0.807 

0.724 

0.62 

0.422 

0.402 

Lagging 

current. 

Ix  —  Eo  sin  d 

0.3 

0.741 

1.021 

1.187 

1.358 

1.635 

2.005 

Where  Eo 

sin  6  is  —  . 

(Ix  -  Eo  sin  5)2 

0.09 

0.55 

1.045 

1.412 

1.845 

2.68 

4.02 

tf« 

1.09 

1.38     . 

1.638 

1.854 

2.153 

2.84 

4.0825 

E 

1.042 

1.172 

1.278 

1.36 

1.466 

1.681 

2.02 

Leading 

current. 

i 

1.0 

1.02 

1.064 

1.113 

1.204 

1.478 

2.362 

ii 

0.0 

±0.411 

±0.745 

±1.005 

±1.34 

±2.013 

±3.22 

mi 

0.5 

0.51 

0.532 

0.556 

0.602 

0.739 

1.181 

-mi)  2 

0.25 

0.26 

0.284 

0.31 

0.363 

0.547 

1.4 

mil 

0.0 

±0.205 

±0.372 

±0.502 

±0.67 

±1.006 

±1.61 

(CE  -  mil) 

1.042 

0.709 

0.435 

0.222 

-0.05 

-0.584 

-1.208 

Lagging. 

(CE  -  mil)  2 

.09 

0.503 

0.19 

0.0494 

0.0025 

0.341 

1.46 

Lagging. 

(CE  -  mil) 

.042 

1.377 

1.650 

1.862 

2.136 

2.687 

3.63 

Leading. 

(CE  -  mil)  2 

.09 

1.9 

2.73 

3.47 

4.56 

7.25 

13.2 

Leading. 

F/2 

.34 

0.763 

0.474 

0.3594 

0.3655 

0.888 

2.86 

Lagging. 

F/« 

.34 

2.16 

3.014 

3.78 

4.923 

7.797 

14.6 

Leading. 

Ff 

.157 

0.873 

0.688 

0.6 

0.605 

0.941 

1.69 

Lagging. 

Ft 

.157 

1.47 

1.735 

1.94 

2.218 

2.788 

3.82 

Leading. 

Case  B: 


I 

1.13 

1.2 

1.4 

1.8 

2.5 

4.0 

Ir 

0.113 

0.12 

0.14 

0.18 

0.25 

0.4 

Ix 

0.339 

0.36 

0.42 

0.54 

0.75 

1.2 

1 

I 

0.885 

0.833 

0.714 

0.555 

0.4 

0.25 

0.1J 

0.113 

0.12 

0.14 

0.18 

0.25 

0.4 

Cos  5 

1.00 

0.953 

0.854 

0.735 

0.65 

0.65 

Ir  —  Eo  COB  6 

-0.887 

-0.833 

-0.714 

-0.555 

-0.4 

-0.25 

(Ir-  Eo  cos  5)2 

0.788 

0.695 

0.51 

0.308 

0.16 

0.0625 

Sin  5 

0.0 

±0.303 

±0.520 

±0.678 

±0.760 

±0.760 

Ix  —  Eo  sin  5 

0.339 

0.057 

-0.10 

-0.138 

-0.01 

0.44 

Where  sin  5 

ia  +  . 

(Ix  -  Eo  sin  5)2 

0.115 

0.00325 

0.01 

0.0191 

0.0001 

0.194 

E* 

0.903 

0.69825 

0.52 

0.3271 

0.1601 

0.2565 

E 

0.95 

0.835 

0.72 

0.572 

0.40 

0.5065 

Lagging 

Ix—  Eo  sin  5 

0.663 

0.94 

1.218 

1.51 

1.96 

Where  5    is 

SYNCHRONOUS  MOTORS 

Case  B:— (Continued) 


339 


(Ix  —  Eo  sin  S)2 

0.44 

0.887 

1.488 

2.285 

3.85 

E* 

1.135 

1.397 

1.796 

2.445 

3.9125 

E 

1.064 

1.18 

1.34 

1.56 

1.977 

Leading. 

i 

1.13 

1.145 

1.196 

1.324 

1.625 

2.6 

mi 

0.565 

0  .  5725 

0.598 

0.662 

0.8125 

1.3 

(-mtV 

0.32 

0.328 

0.359 

0.44 

0.66 

1.69 

ii 

0.0 

±0.3635 

±0.728 

±1.22 

±1.9 

±3.04 

mil 

0.0 

±0.18175 

±0.364 

±0.61 

±0.95 

±1.52 

CE  -  mil 

0.95 

0.653 

0.356 

-0.038 

-0.55 

-1.014 

Lagging. 

(CE  -  mil)  2 

0.903 

0.426 

0.127 

0.001445 

0.303 

1.03 

F/2 

1.223 

0.754 

0.486 

0.441 

0.963 

2.72 

Ff 

1.105 

0.868 

0.697 

0.664 

0.981 

1.648 

Lagging. 

CE  —  mil 

1.246 

1.544 

1.95 

2.51 

3.50 

Treacling. 

(CE  —  mil)2 

1.558 

2.39 

3.8 

6.3 

12.25 

F/2 

1.886 

2.749 

4.24 

6.96 

13.94 

Ff 

1.373 

1.657 

2.057 

2.64 

3.73 

Leading. 

Case  C: 


7 

1.3 

1.35 

1.5 

1.8 

2.5 

4.0 

Ir 

0.13 

0.135 

0.15 

0.18 

0.25 

0.4 

Ix 

0.39 

0.405 

0.45 

0.54 

0.75 

1.2 

1.11 

0.854 

0.823 

0.74 

0.617 

0.444 

0.2775 

0.1117 

0  .  1444 

0.15 

0.1665 

0.2 

0.2775 

0.444 

Cos  S 

1.0 

0.973 

0.9065 

0.817 

0.7215 

0.7215 

Eo  cos  5 

0.9  ' 

0.876 

0.816 

0.735 

0.65 

0.65 

Ir  —  Eo  cos  5 

-0.77 

-0.741 

-0.666 

-0.555 

-0.4 

-0.25 

(Ir-  Eo  cos  S)2 

0.593 

0.55 

0.445 

0.308 

0.16, 

0.0625 

Sin  5 

0.0 

±0.2306 

±0.4222 

±0.5767 

±0.6925 

±0.6925 

Eo  sin  5 

0.0 

±0.2078 

±0.38 

±0.519 

±0.623 

±0.623 

Ix—  Eo  sin  5 

0.39 

0.1972 

0.07 

0.021 

0.127 

0.577 

Where  sin 

8  is  +. 

(Ix  —  Eo  sin  5)2 

0.152 

0.039 

0.0049 

0.00044 

0.0161 

0.333 

E* 

0.745 

0.589 

0.4499 

0.30844 

0.1761 

0.3955 

E 

0.8626 

0.767 

0.67 

0.555 

0.419 

0.629 

Lagging. 

Ix  —  Eo  sin  5 



0.6128 

0.83 

1.059 

1.373 

1.823 

Where     sin 

8  is  -. 

(Ix  —  Eo  sin  5)2 

0.376 

0.69 

1,12 

1.89 

3.33 

#2 

0.926 

1.135 

1.428 

2.05 

3.3925 

S 

0.961 

1.065 

1.193 

1.43 

1.84 

Leading. 

i 

1.3 

1.315 

1.36 

1.472 

1.805 

2.89 

mi 

0.65 

0.6575 

0.68 

0.736 

0.9025 

1.445 

(-mi)2 

0.423 

0.4325 

0.463 

0.542 

0.815 

2.09 

ii 

0.0 

±0.3115 

±0.633 

±  1  .  039 

±1.73 

±2.77 

mti 

0.0 

±0.15575 

±0.3165 

±0.5195 

±0.865 

±1.385 

CE  —  mil 

0.8625 

0.6113 

0.3535 

0.0355 

-0.446 

-0.756 

Lagging. 

(CE  -  mil)2 

0.745 

0.375 

0.125 

0.00126 

0.199 

0.572 

F/2 

1.168 

0.8075 

0.588 

0.54326 

1.014 

2.662 

F/ 

1.08 

0.898 

0.766 

0.736 

1.007 

1.63 

Lagging. 

CE  -  mil 



1.117 

1.382 

1.7125 

2.295 

3.225 

Leading. 

(CE  -  mil)'' 

1.25 

1.915 

2.945 

5.29 

10.4 

F/2 

1  .  6825 

2.378 

3.487 

6.105 

12.49 

Ff 

1.297 

1.54 

1.865 

2.472 

3.535 

Leading. 

340 


ELECTRICAL  ENGINEERING 


7 


3.5 


7 


7 


7 


3.0 


2.5 


7 


2.0 


z 


1.5 


1.0 


0.25  0.5 


0.75  1.0  1.25 

Induced  E.M.F.,  E 

FIG.  251. 


1.5  1.75 


0.5  1.0 


1.5  2.0  2.5 

Field  Excitation,  Ff 
FIG.  252. 


3.0  3.5  4.0 


SYNCHRONOUS  MOTORS 


341 


Phase  characteristic  curves  are  shown  in  Figs.  251  and  252  for 
the  three  cases  considered. 

It  is  also  of  interest  to  see  how  the  phase  characteristic  is  af- 
fected by  a  change  in  the  amount  of  the  motor  load.  Accord- 
ingly curves  are  drawn  for  the  second  case  (EQ  =  1,  E  =  1),  under 


0  0,2  0.4  0.6  0.3  1.0  1.2  1.4  1.6  1.8 

Induced  E.M.F.,  E 

FIG.  253. 

the  three  conditions  P  =  1,  P  =  0.5,  P  =  0.  These  are  shown  in 
Fig.  253.  These  curves  are  sometimes  called  V-curves.  As  the 
friction  loss  is  included  in  the  load,  the  condition  P  =  0  can  never 
be  attained.  In  practice,  the  curve  obtained  with  the  motor 
running  light  approximates  to  this,  however.  The  dotted  line 
gives  the  locus  of  the  minimum  current  points  which  is  also  the 
current  at  unity  power  factor. 


CHAPTER  XLII 
INDUCTION  MOTORS 

The  production  of  torque,  and  the  consequent  operation,  of 
direct-current  motors  is  readily  understood  since  the  condition 
of  wires  carrying  current  placed  in  a  field  at  right  angles  to  the 
direction  of  the  lines  of  force  is  quite  apparent  in  both  the  shunt 
and  the  series  types. 

If  alternating  current  is  supplied  to  the  terminals  of  a  direct- 
current  series  motor,  the  motor  might  reasonably  be  expected 
to  run.  In  such  a  case  the  current  is  the  same  in  both  the  field 
and  the  armature  coils  and  since  the  flux  is  in  time-phase  with 
the  current  which  produces  it,  the  condition  for  the  production 
of  torque  is  satisfied.  Moreover,  since  the  alternation  of  the 
flux  and  the  current  is  simultaneous,  the  direction  of  the  torque 
is  not  changed  though  it  pulsates  in  value.  Such  a  motor  would 
have  a  low  power  factor,  due  to  its  great  inductance,  and  low 
efficiency  due  to  its  great  copper  and  core  losses,  the  latter  being 
excessive  with  unlaminated  field  structure. 

When  alternating  current  is  supplied  to  a  shunt  motor,  the 
condition  for  operation  is  not  so  well  met.  In  this  case,  the  cur- 
rents in  the  armature  and  the  field  coils  will  no  longer  be  in  time- 
phase  with  each  other.  The  current  in  the  field  coils  will  lag  by 
nearly  90  time  degrees  behind  the  voltage,  while  that  in  the 
armature  will  have  only  a  slight  time  lag.  This  difficulty  might 
be  obviated  theoretically  by  placing  a  suitable  condenser  in 
series  in  the  field  circuit.  Practically,  however,  such  a  condenser 
would  be  too  large  and  expensive  to  warrant  its  use. 

The  question  then  naturally  arises:  Why  not  excite  the  field 
from  the  other  phase  of  a  two-phase  supply?  The  trouble  with 
such  a  solution,  assuming  that  a  two-phase  supply  is  available,  is 
that  one-phase  would  be  loaded  with  a  wattless  current ;  moreover, 
the  armature  reaction  and  the  torque  would  be  pulsating. 

A  natural  suggestion  might  be  to  run  two  motors  so  as  to 
balance  the  phases. 

The  next  step  in  the  development  of  the  alternating-current 
motor  would  be  to  omit  the  commutator,  applying  the  well-known 

342 


INDUCTION  MOTORS 


343 


principle  of  the  production  of  currents  by  induction,  as  is  done 
in  the  transformer.  In  this  case  the  field  winding  acts  as  the 
primary,  and  the  armature  winding  as  the  secondary  coil. 
Currents  induced  in  the  armature  would  have  directions  as  shown 
by  the  crosses  and  dots  in  Fig.  254. 

This  arrangement  would  give  no  resultant  torque,  the  torque 
due  to  the  upper  conductors  being  equal  and  opposite  to  that 
due  to  the  lower  conductors. 

Therefore  another  set  of  poles  (shown  dotted)  should  be  intro- 
duced in  space  quadrature  to  the  original  poles,  and  the  flux  due 
to  these  new  poles  should  be  in  time-phase  with  the  armature 
current. 

This  means  a  quadrature  relationship  in  both  time  and  space 
between  the  two  sets  of  poles,  exactly  as  is  the  case  of  a  two-phase 


FIG.  255. 

system.     The  resultant  flux  acting  on  the  armature  forms  the 
well-known  rotary  magnetic  field. 

The  Rotary  Field. — The  production  of  the  rotary  magnetic  field 
may  be  considered  as  due  to  the  currents  in  two  sets  of  coils  as 
shown  in  Fig.  255  (a) .  The  current  in  phase  A  sets  up  an  alternat- 
ing flux  through  the  armature  in  the  horizontal  direction,  while 
that  of  phase  B  sets  up  a  similar  flux  in  the  vertical  direction. 
These  fluxes  have  the  space  relationship  shown  in  Fig.  255  (6). 

The  time  relationship  of  the  fluxes  is  shown  by  their  equations. 
Thus 

<f>A  =  3>m  sin  wt 

fa  =  3>m  sin  (««  +  90) 

The  resultant  flux  at  any  instant  will  then  be  composed  of  a 
horizontal  component  having  the  value  of  <^  at  that  instant, 
and  a  vertical  component  <fo  at  the  same  instant.  This  may  be 
expressed  as 

=  $sin  coJ  -  cos  w 


344  ELECTRICAL  ENGINEERING 

Tabulating: 


M°) 

0 

30 

60 

90 

120 

150 

180 

Sin  wt 

0 

0.5 

0.866 

1 

0.866 

0.5 

0 

Cos  u>t 

1 

0.866 

0.5 

0 

-0.5 

-0.866 

-1 

Sin  ut  —  j  cos  ut 

0  -  jl 

0.5  -  jO.866 

0.866  -  jO.5 

1 

0.866  +  jO.5 

0.5  +  jO.866 

ft 

These  vectors  are  shown  plotted  through  360°  in  Fig. 

Thus,  the  locus  of  the  ends  of  the  resultant  flux  vectors  is  a 
circle  of  radius  <i>w,  and  the  speed  of  rotation  of  the  flux  is  /, 
the  frequency  of  the  alternating  current. 

Problem  108. — In  a  similar  manner,  show  the  relationship  of  the  fluxes 
in  a  three-phase  armature,  and  prove  that  the  resultant  flux  is  a  uniformly 
rotating  vector  of  magnitude  1.5$™. 

Considering,  now,  that  such  a  rotating  flux 
will  be  cutting  the  conductors  of  the  arma- 
ture; if  the  latter  is  short-circuited  it  is  evi- 
dent that  very  large  currents  would  be  induced 
in  it.  As  the  speed  of  the  armature  in- 
creases, the  rate  of  cutting  of  its  conductors 
by  the  flux  decreases,  with  a  consequent 
decrease  of  induced  e.m.f.  and  current.  If 
the  armature  were  to  run  at  synchronous 

speed,  no  current  would  be  set  up  in  its  conductors,  and  hence 

there  could  be  no  torque. 

Theory  of  Operation. — Assuming  a  1:1  ratio  of  turns  of  the 

two  windings,  as  was  done  in  the  case  of  transformers. 


FIG.  256. 


FIG.  257. 

Case  1.  Armature  at  Standstill. — Let  an  electromotive  force 
be  impressed  upon  the  primary,  or  field,  winding  so  as  to  cause 
the  current  70o  to  flow.  This  current  sets  up  a  flux,  </>,  which 
induces  electromotive  forces,  Ei,  in  both  windings.  Since  the 


INDUCTION  MOTORS 


345 


secondary,  or  armature,  is  short-circuited,  the  current  7i  flows, 
so  that  Ei  is  used  up  in  overcoming  the  resistance,  /^i,  and  the 
reactance,  I&i,  of  the  secondary. 

The  primary  current  70  as  in  the  transformer,  must  be 
the  vector  sum  of  70o,  the  exciting  current,  and  —  I\  the  load 
component. 

The  impressed  e.m.f.  E0  is,  likewise,  the  sum  of  the  IoZQ  drop 
and  —  Ei}  that  which  is  supplied  to  overcome  the  induced  e.m.f. 
in  the  primary.  Primary  and  secondary  phase  angles  are  given 
by  00  and  61,  respectively  (Fig.  257). 


Case  2.  Armature  at  about  Halt-speed. — With  the  same  im- 
pressed e.m.f.,  EQ,  the  vector  diagram  for  half-speed  becomes 
altered,  due  to  the  reduced  Ei  in  the  secondary  and  the  reduced 
secondary  reactance.  Assuming  constant  secondary  inductance, 
LI,  the  secondary  reactance,  x\  =  2ir/iLi,  is  directly  proportional 
to  the  difference  in  speed  between  the  rotary  field  and  the  arma- 
ture. This  difference  in  speed,  expressed  in  per  cent,  of  synchro- 
nous speed,  is  called  the  "slip"  of  the  motor,  and  is  denoted  by 
s.  Thus,  at  standstill  s  =  1,  and  x  =  x\]  at  half-speed,  s  =  0.5 
and  x  =  0.5zi.  For  any  speed,  1  —  s,  x  =  sxi. 

As  the  speed  increases,  therefore,  the  secondary  reactance 
becomes  less  important,  0i  decreases,  and  the  value  of  I\  is 
governed  to  a  greater  extent  by  the  secondary  resistance. 

Since  Ei  decreases  in  the  secondary,  7i  also  tends  to  decrease, 
this  tendency  being  counteracted  in  part,  however,  by  the  re- 
duction in  reactance.  The  primary  current,  7o  and  the  IQZQ 
drop  are  reduced  nearly  in  proportion  to  7i.  EQ  being  constant, 
the  voltage  (  —  Ei)  is  somewhat  increased  since  —  Ei  =  EQ  — 
IQZQ.  Therefore  700  is  increased,  and  likewise  the  flux  <f>. 


346  ELECTRICAL  ENGINEERING 

The  entire  induction  motor  circuit  may  be  represented  by  an 
"equivalent  circuit,"  as  was  done  with  the  transformer,  page  178, 
in  which,  however,  the  slip,  s,  enters  as  a  factor  with  reference  to 
both  the  secondary  reactance  and  the  load.  This  diagram, 
Fig.  259,  refers  to  one  phase  only.  EQ  is  phase  voltage,  and 
Ii2R  is  the  nth  part  of  the  motor  load,  where  n  is  the  number  of 
phases.  The  magnitudes  of  the  various  quantities  are  readily 
apparent  from  an  inspection  of  the  "equivalent  circuit"  diagram, 
whatever  may  be  the  load  placed  upon  the  motor. 


FIG.  259. 

Referring  to  the  vector  diagrams,  Figs.  257  and  258,  or  to  the 
"equivalent  circuit"  diagram,  Fig.  259. 

Let 

Im  —  magnetizing  current, 

Ih  =  core-loss  current, 

ZQ  =  TQ  +  j  XQ  =  primary  impedance, 

Zi  =  r\  +  j  sxi  =  secondary  impedance. 
Then 


and 


0oo  =  —  =  conductance  of  exciting  circuit, 


60o  =  —  =  susceptance  of  exciting  circuit, 


where 

et  =  primary  induced  e.m.f. 

and  Im  is  a  positive  quantity  so  that  bQQ  is  always  negative, 

these  quantities  being  all  taken  per  phase. 

Also, 

s  =  slip. 

At  standstill,  s  =  1;  at  synchronous  speed,  s  =  0;  at  normal 
full-load,  s  is  usually  about  0.02  in  per  cent,  of  synchronous 
speed. 

Then, 

S6i  =  secondary  induced  e.m.f. 


INDUCTION  MOTORS  347 

Let  6i  be  chosen  zero  vector.     Secondary  current  may  be  written 


where 

sr1 


1     —          9      I 

T.i  T" 

and 


2      i         9       2 

The  exciting  current  is 

The  primary  current  is 

IQ  =  IQQ  -\-I\-  €i(a\  -\-  00o  +  j(o>z  +  60o)). 


The  e.m.f.   consumed   by  the  primary  impedance  is 
ei(bi  +  jbz)(rQ  -\-jxo)  and  the  impressed  voltage  is 

EQ  =  €i  +  /o20  =  €i  +  et-(6i  +  j62)  (r0  +  jaJo) 

62r0)] 


The  torque,  in  any  motor,  is  proportional  to  the  current  and 
the  flux  in  time-phase  therewith. 

If  1 1  is  the  secondary  current  due  to  a  certain  phase  of  the  pri- 
mary, whose  induced  e.m.f.  is  e^  then  the  power  component  of 
/i  has  been  shown  to  be  e<oi. 

—jl\  is  evidently  the  secondary  current  due  to  a  primary  phase 
which  is  90°  in  space  and  time  behind  the  former.  The  induced 
e.m.f.  of  this  phase  is,  of  course,  —  jei  and  the  flux  causing  the 


348  ELECTRICAL  ENGINEERING 

e.m.f.  is  90°  in  time  ahead  of  the  e.m.f.  Thus  the  flux  which 
reacts  on  the  power  component  of  the  original  secondary  current 
is  proportional  to  j(—  jet)  =  kei. 

.'.  The  torque  is  ke^eiai  —  kefai. 

Torque  is  often  expressed  in  "  synchronous  watts,"  a  term 
which  means  the  number  of  watts  which  would  be  required  to  give 
the  torque  if  the  motor  were  running  at  synchronous  speed. 
k,  then  becomes  1,  and 

T  =  6izai  synchronous  watts  per  phase. 
The  horsepower  per  phase  is 

ef  ai       2irRN  X  lb. 


where 

120  X  frequency 
N  '-=  r'-m'  =          ~- 


and  R  =  radius  of  the  rotor  in  feet. 
Thus  torque  per  phase  is 

T/phase  =  0.059  ««*  X  j-g^  ft.-lb.  ' 

For  a  three-phase  machine 

T  =  3  X  torque  per  phase  =  0.177  efai  X  ?• 

Since  e?a\t  as  Asynchronous  watts"  is  the  output  of  the  motor 
at  synchronous  speed,  then  at  any  other  speed  the  output,  which 
includes  friction  would  be 


Pm  =  d2ai(l  -  s). 

The  power  input  obtained  by  "  telescoping  "  the  vectors  EQ 
and  70  is,  per  phase, 

Po  =  e,-2(d&i  +  c262). 
E  2 

=  rf^^  +  c^)- 

If  fm  =  friction  per  phase,  the  efficiency  is 

_    OUtpUt   _   Pm  —  fm 

input    :         Po 
Power  factor  is 

PQ 


INDUCTION  MOTORS  349 

Apparent  efficiency  is 

Pm-fm 

The  total  output,  neglecting  friction,  is  approximately, 

3  X  et-2ai(l  -  s)  =  — °-^ — ^- 
Ci  ~\~  c<2, 

To  find  the  maximum  output,  this  quantity  may  be  differen- 
tiated and  equated  to  zero.  The  process  is  tedious,  but  by 
neglecting  certain  small  quantities  which  appear,  the  result  may 
be  shown  as  approximately, 


max.  Pm  =  —2 

where 

r  =  rQ  -f-  7*1 

and  

Z  =  V(r0-hri)2+  (x 
If  there  are  p  phases 

max.  Pm  =   pE( 


r+Z 

From  the  equations  just  derived  it  is  possible  to  construct  the 
performance  curves  for  any  motor  for  which  the  constants  are 
given.  These  curves  show  the  efficiency,  power  factor,  apparent 
efficiency,  slip  and  line  current,  all  plotted  against  the  output, 
usually  expressed  in  horsepower. 

Another  set  of  curves  of  great  interest,  particularly  in  respect 
to  the  performance  of  the  motor  at  starting,  consists  of  the  speed 
and  line  current  plotted  against  the  torque. 

In  comparing  actual  motors  by  means  of  performance  curves 
considerable  difficulty  is  encountered  in  determining  relative 
merits.  If  the  curves  are  put  on  the  percentage  basis,  however, 
this  difficulty  vanishes.  In  order  to  study  properly  the  effects  of 
different  variables  in  the  motor  design  a  number  of  typical  cases 
are  worked  out  on  the  percentage  basis,  as  follows: 

Problem  269.— A.  Let  E0  =  1,  r0  =  r:  =  0.02,  x0  =  xi  =  0.12,  Im  =  0.3, 
Ih  =  0.02,  /0  =  0.01  =  friction  loss. 

At  synchronous  speed,  the  exciting  current  only,  exists,  and  EQ  =  Ei 
approximately. 

Therefore  the  constants  g00  and  60o  are  obtained  as: 

300  =  Fo  =  °'02;  6o°   =  ~  fe  =    ~  °'3' 


350 


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INDUCTION  MOTORS 


351 


B.  Same  constants  as  in  A,  except  that  the  magnetizing 
current  is  taken  as  Im  =  0.2.  .'.  60o  =  —  0.2.  A  summary  of 
the  tabulation  is  as  follows: 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.205 

0.55 

0.998 

1.40 

2.09 

2.77 

3.11 

3.96 

4.07 

4.15 

T 

0.0 

0.46 

0.875 

1.2 

1.63 

1.85 

1.82 

1.00 

0.635 

0.33 

P.    -/O 

0.0 

0.445 

0.85 

1.16 

1.54 

1.69 

1.63 

0.69 

0.308 

0.0 

Po 

0.020 

0.485 

0.91 

1.26 

1.73 

2.03 

2.03 

1.325 

0.98 

0.675 

Eff. 

0.0 

0.918 

0.934 

0.92 

0.89 

0.836 

0.8 

0.523 

0.312 

0.0 

P.F. 

0.097 

0.882 

0.908 

0.898 

0.827 

0.722 

0.653 

0.334 

0.241 

0.163 

App.  eff. 

0.0 

0.81 

0.85 

0.825 

0.737 

0.603 

0.523 

0.175 

0.076 

0.0 

C.  Same  as  A,  except  that  resistance  is  inserted  in  the  second- 
ary so  that  7*1  =  0.05. 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.29 

0.36 

0.50 

0.665 

1.01 

1.5 

1.79 

3.36 

3.8 

4.07 

T 

0.0 

0.184 

0.362 

0.532 

0.85 

1.24 

1.44 

1.74 

1.34 

0.77 

Pm    ~/o 

0.0 

0.172 

0.344 

0.505 

0.8 

1.12 

1.29 

1.21 

0.66 

0.0 

Po 

0.0203 

0.205 

0.384 

0.56 

0.881 

1.33 

1.52 

1.96 

1.64 

1.11 

Eff. 

0.0 

0.84 

0.895 

0.902 

0.894 

0.87 

0.845 

0.612 

0.405 

0.0 

P.F. 

0.07 

0.57 

0.77 

0.842 

0.873 

0.866 

0.85 

0.588 

0.431 

0.272 

App.  eff. 

0.0 

0.478 

0.686 

0.76 

0.79 

0.75 

0.718 

0.36 

0.174 

0.0 

D.  Same  as  A,  except  that  the  resistance  of  both  windings  has 
been  increased  so  that  r0  =  r*i  =  0.05. 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.29 

0.358 

0.498 

0.658 

0.82 

1.445 

1.705 

3.15 

3.6 

3.92 

T 

0.0 

0.181 

0.355 

0.519 

0.665 

1.145 

1.315 

1.525 

1.19 

0.714 

Pm    -/O 

0.0 

0.17 

0.338 

0.494 

0.625 

1.043 

1.174 

1.058 

0.586 

0.0 

Po 

0.023 

0.206 

0.386 

0.558 

0.716 

1.307 

1.475 

2.03 

1.85 

1.485 

Eff. 

0.0 

0.825 

0.875 

0.885 

0.872 

0.82 

0.8 

0.52 

0.319 

0.0 

P.F. 

0.0785 

0.575 

0.775 

0.847 

0.874 

0.878 

0.864 

0.645 

0.511 

0.379 

App.  eff. 

0.0 

0.475 

0.678 

0.75 

0.766 

0.72 

0.690 

0.336 

0.163 

0.0 

352 


ELECTRICAL  ENGINEERING 


E.  Same  as  A,  except  that  the  reactance  of  both  windings  has 
been  increased,  so  that  XQ  =  Xi  =  0.18. 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

0.1 

/o 

0.285 

0.58 

0.98 

1.37 

1.86 

2.28 

2.46 

2.78 

2.82 

2.84 

T 

0.0 

0.431 

0.775 

1.01 

1.21 

1.153 

1.057 

0.45 

0.286 

0.144 

Pm  -/o 

0.0 

0.427 

0.75 

0.97 

1.14 

1.052 

0.94 

0.304 

0.133 

0.0 

Po 

0.0196 

0.456 

0.81 

1.06 

1.29 

1.267 

1.173 

0.615 

0.57 

0.308 

Eff. 

0.0 

0.936 

0.926 

0.914 

0.884 

0.832 

0.8 

0.495 

0.298 

0.0 

P.F. 

0.069 

0.786 

0.826 

0.774 

0.692 

0.555 

0.5 

0.218 

0.16 

0.11 

App.  eff. 

0.0 

0.736 

0.765 

0.708 

0.61 

0.461 

0.4 

0.108 

0.047 

0.0 

F.  Same  as  E,  except  that  the  secondary  resistance  is  increased 
so  that  n  =  0.05. 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.285 

0.356 

0.484 

0.66 

1.016 

1.4 

1.64 

2.53 

2.7 

2.83 

T 

0.0 

0.178 

0.354 

0.506 

0.795 

1.04 

1.16 

0.954 

0.651 

0.358 

P»-/o 

0.0 

0.1665 

0.337 

0.481 

0.745 

0.947 

1.03 

0.657 

0.315 

0.0 

Po 

0.0197 

0.1982 

0.377 

0.532 

0.828 

1.09 

1.21 

1.085 

0.802 

0.517 

Eff. 

0.0 

0.84 

0.90 

0.906 

0.9 

0.87 

0.85 

0.606 

0.393 

0.0 

P.F. 

0.069 

0.556 

0.78 

0.81 

0.815 

0.778 

0.737 

0.43 

0.297 

0.185 

App.  eff. 

0.0 

0.47 

0.697 

0.733 

0.734 

0.677 

0.628 

0.26 

0.117 

0.0 

G.  Same  as  A,  except  that  only  half-voltage  is  impressed  on 
the  motor,  i.e.,  E0  =  0.5.     Hence  Im  =  0.15  and  Ih  =  0.01. 


Slip 

0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.145 

0.293 

0.503 

0.71 

1.055 

1.375 

1.565 

2.0 

2.06 

2.085 

T 

0.0 

0.114 

0.211 

0.295 

0.405 

0.456 

0.45 

0.248 

0.158 

0.0814 

P.-/O 

0.0 

0.103 

0.197 

0.276 

0.375 

0.41 

0.395 

0.164 

0.069 

0.0 

Po 

0  .  0051 

0.1195 

0.221 

0.308 

0.428 

0.497 

0.503 

0.329 

0.244 

0.1655 

Eff. 

0.0 

0.86 

0.89 

0.895 

0.875 

0.825 

0.785 

0.5 

0.283 

0.0 

P.F. 

0.07 

0.818 

0.878 

0.869 

0.81 

0.724 

0.644 

0.33 

0.237 

0.16 

App.  eff. 

0.0 

0.705 

0.771 

0.778 

0.709 

0.597 

0.505 

0.174 

0.077 

0.0 

H.  The  same  motor  under  normal  operation  as  in  A.  In  this 
case,  however,  the  secondary  is  so  arranged  as  to  permit  the 
insertion  of  extra  resistances  to  improve  the  torque  while  the 
motor  is  coming  up  to  speed.  From  standstill  to  half-speed 
the  secondary  resistance  is  ri  =  1 ;  from  half-speed  to  nine-tenths 
of  synchronous  speed,  ri  =  0.5;  over  nine-tenths  speed,  r\  =  0.02. 


INDUCTION  MOTORS  353 

This  is  one  way  of  meeting  the  condition  of  starting  under  load. 


n  =  1 

ri  =  0.5 

Slip 

• 

1.0 

0.8 

0.6 

0.5 

0.5 

0.4 

0.2 

0.1 

/o 

1.010.84 

0.67 

0.583 

1.01 

0.84 

0.55 

0.36 

T7 

0.8480.698 

0.536 

0.45 

0.85 

0.70 

0.36 

0.184 

Pm   -/o 

0.0  10.13 

0.204 

0.215 

0.415 

0.41 

0.282 

0.157 

Po 

0.87 

0.727 

0.565 

0.478 

0.87 

0.727 

0.389 

0.206 

Eff. 

0.0 

0.18 

0.36 

0.45 

0.477 

0.564 

0.725 

0.762 

P.F. 

0.865 

0.868 

0.844 

0.815 

0.865 

0.868 

0.77 

0.57 

App.  eff. 

0.0 

0.162 

0.32 

0.38 

0.41 

0.49 

0.57 

0.45 

Performance  curves  are  given  for  all  cases  calculated.  Fig. 
260  shows  for  case  A,  efficiency,  apparent  efficiency,  power  factor, 
current  and  slip,  all  plotted  against  the  output.  The  scales  show 


0.2  0.4  0.6  0.8  1.0 

Output,  (Pm.f0) 

FIG.  260. 


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1.6 


percentage  values.  Similar  sets  of  curves  are  shown  for  the 
other  cases  in  Figs.  261  to  266  inclusive. 

These  curves  illustrate  the  capabilities  of  the  various  motors 
under  normal  running  conditions. 

Of  equally  great  interest  are  the  curves  between  speed,  current 

23 


354 


ELECTRICAL  ENGINEERING 


Induction  Motor  Performance  Curves 

~l,  ro-n-0.02,  £o-£  1-0.12, 
J^-0.02,   JOT"  0.2. 


0.6  0.8  1.0 

Output,  (Pm_/0) 

FIG.  261. 


1.4 


1.6 


3-u| 

8  1.1 


C  Induction  Motor  Performance  Curves 

^o"i.  ?V=o.02f 

1^=0.02,    Jm=0.3. 


1.4 


FIG.  262. 


INDUCTION  MOTORS 


355 


Induction  Motor  Performance  Curves 
#0=1,  Tb-n-O.OS,  »o-S  1-0.12. 
7^=0.02.    /m=0.3. 


0.2 


0.4 


0.6  0.8  1.0 

Output,  (Pm-fo) 

FIG.  263. 


1.2  1.4 


E    Induction  Motor  Performance  Curves 
/&0=0.02,°  1^=0.3. 

Power 


0.4  0.6  0.8 

Output,  (POT-/0) 

FIG.  264. 


356 


ELECTRICAL  ENGINEERING 


1.1 

1.0 


F  Induction  Motor  Performance  Curves 

EQ-=I,  r0=o.o2,  rt=o.c 

It, «  0.02.    /m=0.3. 


0.4  0.6  0.8 

Output,  (Pm.-.f0) 

FIG.  265. 


1.0 


1.2 


G  Induction  Motor  Performance  Curves 

-o.5.  r0=ri=o.o 

-0.01.    /w=0.16. 


0.05        0.1 


0.15        0.2       0.25        0.3 
Output,   (Pm-/0) 

FIG.  266. 


0.35        0.4       0.45 


INDUCTION  MOTORS 


357 


0.2 


0.2  0.4  0.6  0.8 

Speed,  and  Current  -f-  4 

FIG.  267. 


1.2 

1.1 

1.0 

0.9 

0.8 

|0.7 

£0.6 

0.5 

0.4 

0.3 

0.2 

0.1 

0 


7 


\ 


\ 


L 


0.2  0.4  0.6  0.8  1.0 

Speed,  and  Current  -f  2 

FIG.  268. 


1.2 


1.4 


358 


ELECTRICAL  ENGINEERING 


and  torque,  which  are  characteristic  of  starting  conditions.  These 
are  shown  in  Figs.  267,  268,  and  269  for  all  cases  except  B  and 
D.  In  these  two  cases  the  curves  are  very  nearly  the  same 
as  for  case  A.  The  differences  may  be  seen  by  a  glance  at  the 
tabulations. 

Fig.  267  is  of  special  practical  interest.  Here  torque-speed 
curves  are  given  for  a  number  of  motors  which  differ  only  in 
the  amount  of  their  secondary  resistance.  There  is  practically 
no  difference  in  the  current  curves,  one  curve  giving  the  current 
for  all  motors. 


0.5 


0.4 


0.3 


0.2 


0.1 


¥ 


0.2  0.4  0.6  0.8 

Speed,  and  Current  7-  2 

FIG.  269. 


1.0 


A  motor  may  be  imagined  as  supplied  with  a  variable  secondary 
resistance.  Suppose  that  it  starts  with  r\  =  1  ohm.  The 
torque  will  be  0.85,  and  the  current  will  be  0.25  X  4  =  1.00  as  is 
seen  from  the  figure.  Thus,  the  motor  starts  with  full-load 
torque  and  current.  When  half-speed  has  been  attained,  the 
secondary  resistance  is  changed  by  some  device  to  7*1  =  0.5. 
The  motor  at  once  is  changed  from  operation  at  the  point,  a, 
to  the  point,  6,  and  the  current  rises  from  0.6,  to  which  it  had 
fallen,  back  to  the  original  value  of  1.00.  The  motor  now  follows 
the  second  torque  curve  to  c,  then,  by  a  change  of  resistance  to 
7*1  =  0.2,  it  accelerates  along  the  curve  de,  the  current  following 
curve  d'e'.  Another  change  to  TI  —  0.05  causes  the  motor  to  run 
along  fg,  the  current  following  /'</',  at  which  latter  points  the  final 
change  to  TI  =  0.02  is  made. 


INDUCTION  MOTORS  359 

The  motor  now  operates  on  its  normal  running  speed-torque 
curve.  During  all  these  changes  the  current  has  remained  low. 
It  might,  however,  be  more  desirable  to  utilize  the  variation  in 
the  secondary  resistance  to  maintain  a  very  high  torque  from 
starting.  In  such  a  case,  the  start  would  be  made  with  TI  =  0.2. 
At  half-speed  a  change  to,  say,  ri  =  0.1  would  be  made.  In 
this  way,  approximately  double  full-load  torque  could  be  main- 
tained during  the  accelerating  period,  the  current,  however,  being 
correspondingly  heavy. 

Example.  —  As  a  particular  example  to  illustrate  the  use  of  the 
preceding  figures,  consider  a  three-phase,  6-pole,  60-cycle  motor 
of  75  hp.  and  440  volts. 

Let  its  constants,  in  percentage,  be  those  of  case  A,  and  let 
it  be  required  to  find  the  following: 

1.  Full-load  current. 

2.  Starting  current. 

3.  Starting  torque. 

4.  Impressed  voltage  to  give  normal  current  at  starting. 

5.  Maximum  output  in  horsepower. 

6.  Starting  torque  with  normal  current. 

7.  Maximum  output  under  voltage  required  to  give  normal 
current  at  starting. 

(a)  Motor  Assumed  Y-connected.  —  Since,  in  the  curves,  out- 
put is  expressed  in  watts  per  phase,  the  output  of  the  motor 
considered  becomes,  at  full-load, 

75 
Pm  -  /o  =  -o-  X  746  =  18,700  watts  per  phase. 

o 

Since  full-load  has  been  taken  as  occurring  at  2  per  cent,  slip, 
this  occurs  also  at  (Pm  -  /„)  =  0.83  in  Fig.  260.  Therefore 
unity  on  the  output  scale  is,  for  the  machine  in  question, 

1  8  700 

'  0     =  22,500.     This  gives  the  output  scale. 

U.oo 

All  the  ordinates,  except  current,  i.e.,  efficiency,  apparent  effi- 
ciency, power  factor,  and  slip,  remain  as  in  the  figure,  being 
correct  for  any  size  motor  of  the  constants,  in  percentage,  of  this 
particular  case. 

18  700 
1.  Full-load   current  =      '  Q    -5-  apparent  efficiency  at  full- 


1  8  700 
load  -  -  90'3 


360  ELECTRICAL  ENGINEERING 

To  find  the '  current  at  any  other  load,  the  reading  on  the 

90  3 
current    curve   should    be   multiplied    by    — ^  =  88.5.,  where 

1.02  is  the  full-load  current  as  read  on  the  curve. 

2.  Starting  current  is,  therefore,  4.19  X  88.5  =  371  amp.  per 
phase. 

3.  The  torque  expressed  in  synchronous  watts  per  phase,  is 
shown  in  Fig.  267.     Expressed  in  ft.-lb.  at  the  pulley  at  synchro- 
nous speed  the  torque  becomes,  torque  =  " synchronous  watts" 

X  0.059  X  number  of  phases  X       P°  es       =  0.177et2ai  -f  for 

frequency  J 

three-phase  motors. 

In  the  example,  p  =  6,  /  =  60. 
Full-load  torque,  at  synchronous  speed,  is 
_  75X33,000  _ 

2*  X  1200 

Therefore,  in  synchronous  watts,  this  is,  328  =  0.177^1  X 
0.1,  or 

328 
T  =  efdi  =  Q  0177  =  18,520  synchronous  watts. 

As  the  curve  (ri  =  0.02,  Fig.  267)  gives  full-load  torque  =  0.856 
synchronous  watts,  to  obtain  values  for  the  particular  machine 

being  considered,  the  torque  at  any  point  on  the  curve  should  be 
1  o  Kon 

multiplied  by  A  '    „    =  21,650,  to  give  synchronous  watts,  or  by 

U.ooo 
ooo 

— — —  =  383  to  give  foot-pounds  at  synchronous  speed. 

U.oOD 

Starting  torque  is,  then,  0.322  X  21,650  =  6970  synchronous 
watts,  or  0.322  X  382  =  123.3  ft.-lbs. 

4.  Having  already  found  the  normal  and  starting  currents, 
and  since  the  current  is  proportional  to  the  voltage,  the  impressed 
voltage  necessary  to  give  normal  current  at  starting  is:  normal 

,  normal  current          440  vx  90.3  . 

phase  voltage  X  -r — r- ~i>  or  — ?=  X  -^n~  -  61.8  volts  to 

starting  current        -\/3       371 

neutral,  or  107  volts  between  terminals. 

5.  Maximum  output  is  (Fig.  260)    1.67  watts.     This   corre- 
sponds to  a  maximum  torque  of  1.82  synchronous  watts  at  8  per 
cent,  slip  (Fig.  267)  these  values  being,  of  course,  per  phase.     To 
change  to  horsepower,  this  gives 

1   f\7 

max.  hp.  =  -~r  =  0.00224  per  phase. 


INDUCTION  MOTORS  361 

For  the  motor  considered, 

max.  output 

max.  hp.  =  -      —  ;  —         -  X  75  hp. 
normal  output 

=  g|g  X  75  =  150  hp. 

6.  Starting  torque,  if  normal  current  only  is  allowed,  is,  since 
torque  is  proportional  to  (voltage)2, 

/ft  -«    o\  2 

T  =  f(y  '  N2  X  starting  torque  at  normal  voltage. 


=  0.0592  X  6970  =  412.5  synchronous  watts,  or 
=  0.0592  X  123.3  =  7.3  ft.-lb. 

7.  Maximum  output  under  impressed  voltage  of  61.8  volts  per 
phase  is  equal  to  maximum  output  under  normal  voltage,  mul- 

tiplied by  the  factor   r^f)    =  0.0592,  or  maximum  output  at 

61.8  volts  =  150  hp.  X  0.0592  =  8.88  hp. 

(b)  Motor  Assumed  A-connected.  —  The  student  may  show  that 
in  this  case  the  required  values  are: 

1.  Full-load  current  =  90.3  amp. 

Phase  current  =  52.2  amp. 

2.  Starting  current  =  214  amp.  per  phase. 

3.  Starting  torque  =  123.3  ft.-lb. 

4.  Voltage  to  give  normal  current  at  starting  =  107. 

5.  Maximum    output  =  150    hp. 

6.  Starting  torque,  with  normal  current,  =  7.3  ft.-lb. 

7.  Maximum  output  on  107  volts  =  8.88  hp. 

Questions.  —  Do  these  answers  indicate  that  a  motor  built  to 
be  operated  Y-connected,  may  be  reconnected  A  and  will  then 
give  substantially  the  same  performance? 

Show  that  if  a  certain  motor  A-connected  is  intended  for 
operation  at  100  volts,  if  it  be  reconnected  Y,  and  operated  at 
the  same  voltage,  the  output  will  be  reduced  to  J^.  Discuss, 
also,  the  effect  of  this  change  on  the  power  factor,  maximum 
output,  torque,  etc. 

Show  that  if  the  impressed  voltage  is  reduced  by  10  per  cent., 
the  maximum  output  and  the  starting  torque  are  reduced  by 
about  20  per  cent.,  the  starting  current  by  about  10  per  cent., 
while  the  efficiency  remains  about  the  same  and  power  factor  is 
slightly  improved  at  light  loads.  This  question  is  very  practical 


362  ELECTRICAL  ENGINEERING 

since  a  motor  designed  for  and  rated  at  125  volts  may  often  be 
available  for  operation  on  a  110- volt  circuit. 

Show  that  such  operation  might  not  always  be  practical  on 
account  of  the  reduction  in  the  overload  range. 

Would  it  be  practical  to  operate  at  10  per  cent,  above  normal 
voltage?  Show  that  in  this  case  the  power  factor  will  be  much 
poorer,  particularly  at  light  loads. 

Show  that  if  the  primary  and  secondary  reactance  are  eaeh 
10  per  cent.,  the  starting  current  will  be  slightly  less  than  five 
times  the  full-load  current. 

From  the  curves  that  have  been  given,  discuss  the  value  of 

starting  current 

the  ratio,   : r  ,  , ?•     When   considering  the  per- 

'   running  light  current 

formance  of  the  motor,  especially  the  margin  in  output,  show 
that  the  smallest  ratio  for  a  good  motor  should  be  about  12. 
The  running  light  current  is  substantially  equal  to  the  mag- 
netizing current. 

Many  other  considerations  are  involved  in  the  choice  or 
operation  of  induction  motors.  Some  of  these  may  be  briefly 
discussed. 

It  has  been  shown  that  a  variation  of  the  secondary  (usually 
the  rotor)  resistance  is  accompanied  by  marked  changes  in  the 
performance  characteristics  of  the  motor;  that  higher  resistance 
means,  roughly,  increased  starting  torque  but  decreased  normal 
running  efficiency.  For  normal  operation,  therefore,  the  smallest 
possible  secondary  resistance  is  desirable.  This  is  best  obtained 
by  a  type  of  rotor  winding  construction  known  as  the  "squirrel 
cage."  In  this,  the  conductors  are  heavy  copper  bars,  lightly 
insulated,  with  only  one  bar  to  a  slot. 

The  ends  of  the  bars  are  connected  to  copper  rings  which 
thus  give  a  completely  short-circuited  winding.  The  resistance 
of  such  a  squirrel  cage  affair  is  extremely  low,  and  the  starting 
torque  of  the  motor  is  correspondingly  low.  There  is  no  op- 
portunity of  inserting  additional  resistance  in  such  a  structure. 
For  this  reason,  many  rotors  are  supplied  with  definite  windings 
the  terminals  of  which  may  either  be  brought  out  to  slip  rings  on 
the  shaft,  or  be  connected  to  a  revolving  resistance  mechanism 
carried  within  the  rotor  spider. 

The  type  of  motor  to  be  chosen  depends  on  the  use  to  which 
it  is  to  be  put.  Induction  motors  cannot  be  used  as  indiscrimi- 
nately as  can  direct-current  motors,  for  example.  Consider  a 


INDUCTION  MOTORS  363 

motor  to  be  used  in  pumping  against  a  high  hydraulic  head.  The 
squirrel  cage  motor  would  not  start.  There  might  also  be  diffi- 
culty with  the  wound  rotor  type.  In  this  case  it  would  be 
necessary  to  have  many  steps  in  the  secondary  rheostat  to  insure 
against  the  torque  falling  at  any  instant  below  the  required 
amount. 

Another  question  of  importance  relates  to  the  frequency. 
Assume,  for  example,  that  a  5-hp.,  60-cycle,  220-volt  motor  is 
required.  If  it  is  found  that  there  is  a  5-hp.,  40-cycle,  220-volt 
motor  available,  will  it  be  practicable  to  utilize  this  machine, 
thus  saving,  perhaps,  the  cost  of  a  new  motor? 

When  a  40-cycle  motor  is  operated  on  the  60-cycle  circuit  it  is 
evident  that  the  magnetizing  current,  and  consequently  the  flux, 

40 

will  be  reduced  in  amount  approximately  in  the  ratio  ^~,  while 

OU 

the  reactance  will  be  correspondingly  increased.  The  motor 
would  therefore  be  weak  in  operation,  that  is,  it  may  have 
insufficient  margin  in  overload  range. 

Changing  from  60  cycles  to  40  cycles  would  have  just  the 
reverse  effect.  The  motor  would  have  ample  capacity.  Mag- 
netic densities  might  be  excessive,  and  the  power  factor  might  be 
considerably  poorer  owing  to  the  great  increase  in  magnetizing 
current. 

To  operate  the  40-cycle  motor  on  60  cycles  would  be  most 

60 
satisfactory  jf  the  voltage  could  be  increased  in  the  ratio  -^ 

This,  however,  is  ordinarily  impossible.  It  is  sometimes  pos- 
sible, however,  to  accomplish  approximately  the  same  result  by 
reconnecting  the  windings.  Suppose,  for  instance,  that  the 
motor  is  A-connected.  Consider  changing  to  Y-connection,  at 
the  same  time  dividing  each  phase  into  two  circuits  and  connect- 
ing them  in  parallel.  Connecting  in  parallel  changes  the  required 
voltage  from  220  to  110.  Changing  to  Y  makes  the  required 
voltage  1.73  X  110  =  190.  The  change  of  frequency  alone 

60 

would  suggest  a  voltage  of  110  X  TQ  =  165.  Perhaps,  there- 
fore, the  change  to  the  condition  of  190  volts,  that  is  to  parallel 
Y-connection,  will  give  satisfactory  results  in  operation. 

An  approximate  value  for  the  power  factor  may  be  obtained 
as  indicated  in  Fig.  270.  A  right  triangle  is  formed,  the  per- 
pendicular sides  of  which  are  per  cent,  load  and  per  cent.  Im 


364  ELECTRICAL  ENGINEERING 

+  per  cent,  reactance.  The  reactance  is  the  sum  of  both  the 
primary  and  the  secondary  reactance  in  per  cent.  The  angle 
00  is  the  phase  angle.  The  basis  for  this  approximation  is  found 
from  a  study  of  the  vector  diagrams,  Fig.  257. 
Motor  and  Transmission  Line.  —  When  an  induction  motor  is 
^  at  the  end  of  a  transmission  line  on  which 

^^^^  constant  voltage  is  impressed,  the  constants 


*  %x      _iL_±^  of  the  line  should  be  added  to  those  of  the 
%Load  motor  windings  in  determining  the   per- 

FIG.  270.  formance  characteristics.     Thus,  r0  is  the 

sum  of  the  primary  winding  and  the  line  resistances,  and  XQ  is 
the  sum  of  the  primary  winding  and  the  line  reactances. 

Let  it  be  assumed  that  a  220-volt  motor  is  at  the  end  of  a 
transmission  line  such  that  maximum  output  occurs  when  the  line 
drop  has  reduced  the  voltage  on  the  motor  to  190. 

/190\  2 
The  maximum  output  is  then  (oon)    =0.75  times  its  value 

under  normal  voltage. 

If  the  performance  curves  at  normal  voltage  are  given,  these 
may  be  changed  to  give  approximately  the  performance  under 
the  conditions  named  by  merely  altering  the  scale  of  abscissae 
so  that  the  maximum  output  shall  occur  at  three-quarters  of  its 
former  value. 

Motor  with  Auto-transformer.  —  If  a  motor  is  used  where  it 
does  not  have  to  start  under  heavy  load,  an  auto-transformer 
may  be  introduced  to  reduce  the  starting  current. 

For  instance,  if  the  auto-transformer  supplies  half  voltage 
the  current  will  be  reduced  to  one-half,  and  the  volt-amperes  will 
consequently  be  only  one-quarter  of  normal  starting  amount. 
On  the  primary  side  of  the  auto-transformer,  then,  the  current 
input  will  be  only  one-quarter  of  the  normal  starting  current. 
Such  an  arrangement  is  advantageous  from  the  standpoint  of 
current,  but  is  bad  where  a  large  starting  torque  is  required,  since 
the  torque  is  reduced  to  one-quarter  its  normal  starting  value. 

Many  other  considerations  may  naturally  arise  in  reference 
to  the  induction  motor,  some  of  which  may  well  be  studied  from 
the  standpoint  of  its  design  which  is  taken  up  in  the  next  chapter. 


CHAPTER  XLIII 

STUDY  OF  THE  DESIGN  CONSTANTS  OF  AN  INDUCTION 

MOTOR1 

The  primary  limiting  features  in  induction  motor  design 
are  (1)  the  possible  amount  of  copper  per  inch  of  periphery,  and 
(2)  the  smallness  of  the  air  gap.  A  small  air  gap  means  a  cheap 
motor  and  high  efficiency.  Gap  lengths  ordinarily  run  from 
0.02  in.  to  0.06  in.,  the  smaller  values  being  for  small  machines. 
A  relation  between  these  two  limiting  features  may  be  found  in 
practice;  thus,  for  each  0.01  in.  of  air  gap,  from  100  to  150  amp.- 
conductors  per  inch  of  periphery  may  be  assumed  as  a  starting 
basis. 

Let  it  be  required  to  design  the  following  motor: 

7-6-10-1200-1 lOv. 

Since  practically  all  induction  motors  are  three-phase,  that 
feature  is  not  indicated  in  the  rating.  If  the  motor  were  two- 
phase,  its  designation  would  be  IQ.  Otherwise,  the  rating 
indicates  6  poles,  10  hp.,  1200  r.p.m.  at  synchronous  speed,  and 
110  volts. 

Air  Gap. — The  air  gap  may  first  be  assumed  in  accordance  with 
practical  experience.  As  a  safe  average  value,  let  this  be  chosen 
as  0.02  in.  =  lg.  The  number  of  ampere-conductors  per  inch 
of  periphery,  from  the  relation  given  above,  will  then  be 

2  X  (100  to  150)  =  200  to  300. 

Rotor  Diameter. — This,  also,  may  be  taken  from  experience. 
As  a  trial  value,  let  the  diameter  per  pole  be  assumed  as  2.5  in. 
The  rotor  diameter  is  then  2.5  X  6  =  15  in.  =  d*. 

Trdi       15.04  X  3.14 

Pole  pitch  =  -«-  = ^ =  7.88  in., 

o  o 

where  di  is  the  inside  diam.  of  the  stator. 

1  It  is  realized  of  course  that  since  this  is  not  a  book  on  design  of  electric 
machines,  the  equations  used  are  not  claimed  to  be  accurate,  but  they  are 
sufficiently  accurate  for  the  purpose  in  view  which  is  to  give  a  fairly  com- 
plete idea  of  how  the  various  constants  are  derived.  Indeed  the  methods 
suggested  have  been  used  in  the  design  of  a  large  number  of  machines  that 
have  been  built. 

365 


366  ELECTRICAL  ENGINEERING 

Stator  Slots  per  Pole. — This  depends  primarily  on  the  slot 
pitch,  but  must  be  a  multiple  of  three.  For  low-voltage  motors 
the  slot  pitch  may  be  quite  small,  say  0.65  in.  As  the  voltage 
increases  the  space  requirements  of  insulation  will  cause  an 
increase  in  the  pitch.  Assuming  0.65  in.,  the  number  of  stator 
slots  per  pole  is 


0.65 
and  the  slot  pitch  revised  is 

7.88  _ 
12 

Slots  per  pole  and  phase  are 


0.657  in. 


- 

:  3 

Ampere-conductors  per  slot  will  be 

(200  to  300)  X  0.657  =  130  to  200. 

The  number  of  conductors  per  slot  will  depend  for  their  size 
on  whether  the  motor  is  to  be  A-  or  Y-connected.  The  current 
they  must  carry  may  be  calculated  on  the  basis  of  10  kv.a. 
input,  or  3300  volt-amp,  per  phase. 

If  A-wound,' 

T  330° 

i  phase    = 


=  \/3  X  30  =  52  amp. 

amp.  conds. 
slot  =  —  -  -  = 
amp. 

If  Y-wound, 


amp.  conds.       130  to  200 

Conductors  per  slot  =  —  -  -  =  -  ^  -  =  4.3  to  6.7 

amp.  60 


Iphase   =   Iline   =   52  amp. 

Conductors  per  slot  =  -  =~  -  =  2.5  to  3.8. 

oz 

Slot  and  Tooth  Dimensions.  —  In  general,  it  is  good  practice  to 
use  if  possible  four  coils  per  slot.  This  arrangement  lends  itself 
readily  to  reconnection  either  in  series-parallel,  for  double  vol- 
tage, or  in  series  for  quadruple  voltage.  Therefore,  selecting  four 
as  the  number  of  groups  of  wires  per  slot  we  also  get  four  effect- 
ive conductors  per  slot.  The  next  step  is  to  design  a  suitable 
slot.  The  deeper  the  slot,  the  more  copper  per  inch  of  periphery 
is  possible. 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       367 


In  a  given  coil,  however,  it  is  not  practical  to  wind  more  than 
four  wires  on  edge.  Therefore,  a  slot  similar  to  that  shown  in 
Fig.  271  should  be  chosen.  This  gives  four  wires  per  coil,  four 
effective  conductors  per  slot,  or  a  total  of  sixteen  wires  per  slot. 
Either  A-  or  Y-connection  may  be  chosen.  In  this  case  it  will 
be  the  latter.  The  actual  slot  dimensions  for  this  motor  are 
depth  =  1.15  in.,  width  =  0.34  in. 
Each  effective  conductor  consists  of 
four  No.  10  B.  &  S.  double  cotton- 
covered  wires.  Each  wire  has  a 
diameter  of  0.112  in.  over  insulation, 
and  the  copper  area  is  0.00815  sq.  in. 

52 

Current   density  is  43^0815  = 

1600  amp.  per  sq.  in.,  which  is  a  rea- 
sonable value. 

The  slot  insulation  is  about  0.04  in. 
in  thickness,  being  sufficient  for  440 
volts. 


U—o— -4^.22 


FIG.  271. 


The  slot  opening  is  made  as  small  as  possible  to  permit  con- 
venient insertion  of  the  coils.  Its  width  in  this  case  is  0.22  in., 
or  about  two-thirds  of  the  slot  width. 

The  tooth  dimensions  are:  width  at  face  =  0.43  in.,  width  at 
narrowest  point  =  0.324  in.,  shown  respectively  at  a  and  b  in  the 
figure. 

Main  Flux. — All  data  are  now  available  to  substitute  into  the 
fundamental  equation 

E  = 


108 
from  which  the  required  flux,  $,  may  be  ascertained: 

E  =  —j=.  =  63.5  volts  to  neutral, 
V3 

/  =  60  cycles, 

t  =  2  turns  per  slot  X  24  slots  per  phase  =  48  turns, 

k  =  constant  for  winding  distribution  =  0.96. 

.  63.5  X  108  ,lfinon 

'  '  *  :=  4.44  X  60  X  48  X  0.96  =     L6'°( 

Stator  Length. — The  cross-sectional  area  of  the  iron  necessary 
to  accommodate  this  flux  is  limited  by  the  maximum  permissible 
flux  density  in  the  stator  teeth.  For  60-cycle  motors  the  maxi- 


368  ELECTRICAL  ENGINEERING 

mum  density  should  not  exceed  90,000  lines  per  sq.  in.;  for  25 
cycles  the  density  may  reach  105,000.  Higher  densities  are 
apt  to  cause  excessive  heating  of  the  teeth. 

It  has  been  shown  that  although  the  windings  on  the  stator 
are  stationary,  the  effect  due  to  the  multiphase  currents  in  them 
is  similar  to  that  of  a  revolving  field  structure.  With  alternators, 
the  field  flux  is  more  or  less  evenly  distributed  along  the  surface 
of  the  pole,  that  is,  the  density  is  fairly  uniform,  and  thus  the 
flux,  when  plotted,  approaches  rectangular  shape.  With  the 
distributed  windings  of  multiphase  induction  motors,  the  space 
distribution  of  the  flux  is  practically  that  of  a  sine  wave,  as  will 
be  seen  by  plotting  the  magnetomotive  forces  at  different  points 
along  the  stator  periphery.  With  a  sinusoidal  space  distribution, 

then,  the  maximum  flux  density  is  ~  times  the  average  density 

over  the  surface. 

The  net  stator  length  may  thus  be  determined  by  assuming 
maximum  teeth  density,  from  the  relation, 
N  ,        ,    __  flux  per  pole  _ 

~~  max.  flux  density  X  min.  teeth  width  per  pole 
This  equation  may  be  used,  however,  more  advantageously  as  a 
check  in  settling  the  final  dimensions  both  of  the  teeth  and  the 
stator  length.  The  latter  may  best  be  determined  at  the  start, 
by  assuming  a  value  for  magnetizing  current  and  working  through 
the  gap. 

Let  the  magnetizing  current  required  to  send  flux  through  the 
gap  be  assumed  as  20  per  cent,  of  full-load  current,  =  0.2  X  52 
=  10.4  amp. 

Gap    amp.-turns  =  \/2  X  1.5  X  10.4  X  8  =  176.5,  since  there 
are  eight  turns  per  pole  and  phase. 
Average  gap  flux  density  is 

D  176-5        _  O 

&avg.  gap   -    Q  ^3   x  Q  Q2  - 

.'.  Gap  area  per  pole, 

516,000 

=  18'3  S* 


teeth  width  per  pole  =  (0.43  +  0.02)  X  12  =  5.4  in.;  where  0.02 
in.  has  been  added  to  the  tooth  width  on  account  of  unavoidable 
irregularities  in  the  stampings  or  laminations. 

18  3 
.*.  gross  stator  length  =  -=-~-  =  3.4  in. 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       369 

Experience  in  design  would  now  lead  one  to  judge  this  length 
of  3.4  in.  to  be  too  short,  as  compared  with  a  rotor  diameter  of 
15  in.  Therefore  it  will  be  advisable  to  repeat  the  calculations 
on  the  basis  of,  say,  12  in.  rotor  diameter,  instead  of  15  in. 

The  new  slot  pitch  then  becomes, 

0.657  X  ~  =  0.525  in. 
10 

The  slot  and  the  slot  opening  remain  as  before,  but  the  tooth 
face  is  reduced  to 

0.525  -  0.22  =  0.305  in. 
Teeth  width  per  pole  =  (0.305  +  0.02)12  =  3.9  in. 

I  Q    O 

.'.  gross  stator  length  =  -~r  =  4.7  in. 

o.y 

and   effective   stator  length  =  0.9  X  4.7  =  4.23  in. 

This  value  may  now  be  applied  to  the  stator  teeth  in  order  to 
ascertain  whether  the  maximum  flux  density  is  satisfactory  or  not. 

Minimum  width  of  tooth  =  0.19  in. 

Minimum  net  area  of  teeth  per  pole  =  0.19  X  12  X  4.23  — 
9.65  in. 

p  -i  n  (\f\r\ 

Therefore,  maximum  average  flux  density  in  teeth  =     n  ' 

y.oo 

=  53,500  lines  per  sq.  in.,  and  maximum  flux  density  in  teeth 
=  I  X  53,500  =  84,000. 

Rotor  Slots. — Consider,  first,  the  squirrel  cage  rotor.  In 
choosing  the  number  of  slots  it  is  well  to  insurq  that  there  shall 
be  no  possible  symmetrical  arrangement  of  the  stator  and  rotor 
slots  with  respect  to  each  other.  In  practice  it  is  common 
to  take  one-half  the  stator  slots  ±  1. 

72 
In  this  case  the  number  will  be  -^  -f  1  =  37  slots.     Such 

a  choice  is  made  to  prevent  the  existence  of  so-called  "dead 
points,"  or  positions  of  the  rotor  from  which  it  may  not  start. 
Slot  and  Tooth  Dimensions. — At  synchronous  or  operating  speed 
the  frequency  of  reversal  of  the  flux  in  the  secondary  core  is 
so  low,  being  that  of  the  slip,  that  the  core  loss  in  the  secondary 
is  negligible  with  proper  teeth  dimensions.  This  fact  permits 
higher  densities  in  the  teeth  than  are  permissible  in  'the  primary 
teeth.  The  maximum  density  occurs,  of  course,  at  the  base  of 

24 


370 


ELECTRICAL  ENGINEERING 


the  teeth.  When  the  rotor  diameter  is  not  great,  as  in  small 
motors,  a  deep  slot  means  considerable  reduction  of  tooth  area  at 
the  base  compared  with  the  area  at  the  top  of  the  tooth.  Usually 
rotor  slots  are  fairly  shallow.  The  dimensions  in  the  present 

case  are  given  in  Fig.  272.  The  rotor 
bar  is  nearly  square  in  section,  and  is 
inserted  in  the  slot  from  the  end.  It 
is  covered  with  a  thin  layer  of  paper 
insulation,  although,  since  the  bars 
are  all  short-circuited,  even  this  is  not 
essential.  The  opening  between  the 
teeth  is  narrow  in  order  to  give  a 
large  flux  area  in  the  gap.  At  the  same  time,  it  is  maintained 
for  the  purpose  of  reducing  the  leakage  flux  and  thereby  cutting 
down  the  self-inductive  reactance. 

Considering  the  dimensions,  as  given,  the  minimum  net  area 
of  rotor  teeth,  per  pole,  is 
07 
~  X  0.32  X  4.23  =  8.35  sq.  in. 

4 

.'.  Average  maximum  flux  density  in  rotor  teeth  is 


516,000 
8.35 


=  61,800  lines  per  sq.  in. 


Maximum  density  =  ^  X  61,800  =  97,000. 

If,  instead  of  the  squirrel  cage,  a  wound  rotor  is  desired,  the 
number  of  slots  chosen  will  in  this  case 

Mi 


be  54,  which  will  give  nine  slots  per 
pole,  and  three  slots  per  pole  and  phase. 

The  rotor  need  not  necessarily  have 
a  three-phase  winding;  any  multiphase 
winding  with  the  proper  number  of 
poles  will  do.  In  fact,  the  squirrel 
cage  may  be  regarded  as  a  winding  of  many  phases. 

Slot  and  tooth  dimensions  are  given  in  Fig.  273.     From  these, 
minimum  teeth  area  per  pole  is  0.21  X  9  X  4.23  =  8  sq.  in. 

.'.  Average  flux  density  in  the  rotor  teeth  at  base  is  — — 


FIG.  273. 


64,500  lines  per  sq.  in.     Maximum  density  =  9  X  64,500  =  101,300. 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       371 

The  remaining  rotor  calculations  are  made  in  essentially  the 
same  way  for  both  the  squirrel  cage  and  the  wound  rotor  types. 

The  former  type  will  therefore,  alone,  be  worked  out.  The 
student  may,  for  practice,  apply  the  process  to  the  determination 
of  constants  and  performance  characteristics  of  the  motor  with 
wound  rotor. 

Rotor  Secondary  Resistance. — This  is  difficult  to  calculate  by 
the  usual  method  applied  to  definite  windings,  but  may  be  deduced 
from  a  well-known  fact  that  the  loss,  per  cubic  inch  of  copper, 
is  0.79  watt  at  60°C.  when  the  current  density  is  1000  amp.  per 
sq.  in.,  and  this  loss  varies  as  the  square  of  the  current  density. 

The  problem,  then,  is  to  find  the  volume  of  the  rotor  con- 
ductors and  the  current  in  them.  The  latter  may  be  considered 
first. 

Consider  the  revolving  flux  as  represented  at  some  given 
instant  by  a  pair  of  poles,  shown  in  Fig.  274.     At  that  instant 
the  current  flows  downward  through  the 
rotor  bars  under  one  pole,  and  upward 
through  the  bars,  under  the  other  pole. 
These    currents    are    indicated    by   "the 
crosses    and    dots  in  the  figure.      Maxi- 
mum current  will  be  in  the  bars  under 
the  middle  of  each  pole  where  the  flux  is 
maximum.     Between  the  poles  there  will  -pio.  274. 

be  little  current  in  the  bars.     In  general, 

the  distribution  of  current  in  the  bars  around  the  periphery 
will  be  proportional  to  the  sine  of  the  space  angle  between  the 
bar  and  the  neutral  axis. 

Since,  now,  there  is  an  inductive  relationship  between  the 
primary  and  the  secondary  windings,  as  in  the  transformer,  if  the 
exciting  current  is  neglected  and  perfect  mutual  inductance  is 
assumed,  the  ampere-conductors  around  the  stator  periphery 
must  equal  the  ampere-conductors  around  the  rotor  periphery. 
But  the  former,  denoted  by 

IaC3  =  288  X  52  =  15,000, 

since  there  are  288  conductors  on  the  stator,  each  carrying 
52  eff.  amp. 

The  effective  rotor  current  per  bar  will  then  be 

15,000 
IT  =  =  405  amp. 


372 


ELECTRICAL  ENGINEERING 


Maximum  current  in  any  bar  is  then, 

7r(max.)  =  405  X  A/2  =  572  amp. 

2 
Average  current  per  bar  =  7r(av.)  =  572  X  -  =  364  amp. 

In  the  end  rings  which  short-circuit  the  bars  the  current  will 
vary  in  amount  along  the  circumference.  The  section  of  the 
ring  at  A,  Fig.  275,  will  carry  the  most  current,  while  at  B,  the 
current  will  be  zero.  One-quarter  of  the  bars  will  send  their 
currents  through  section  A,  or,  in  general,  section  A  will  contain 

total  bars 
the  current  from 


37 


2  X  poles 
In  this  case,  then,  there  will  be  j7>  =  3.08  bars    supplying 

this  current.     The  current  through  A  will  be  3.08  times  the 
average  current  per  bar,  or, 

3.08  X  364  =  1122  amp. 

The  average  current  around  the  whole  circumference  of  the  ring 
is 

1122  X  -  =  715  amp. 

and  the  effective  current  in  the  ring  is 
IR  =  1122  -T-  A/2  =  794  amp. 
Volume  of  each  rotor  bar  is 

0.5  X  0.55  X  6.7  =  1.84  cu.  in. 
Total  volume  of  bars  is 

1.84  X  37  =  68.1  cu.  in. 


FIG 


Since  effective  current  per  bar  is  405  amp.,  and,  assuming  0.79 
watt  per  cu.  in.  loss  at  1000  amp.  per  sq.  in.,  the  loss  in  the  bars 
is 

X  6!U  =  11 


°-79  X  (o.5  X  045°55 


Loss  in  bars  per  phase  =  -5-  =  39  watts. 

o 

Volume  of  end  rings  is 

2  X  0.5  X  TT  X  10.1  =  31.75  cu.  in. 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       373 

Effective  current  per  ring  is  794  amp. 
.'.  Loss  in  the  rings  is 

794       \  2 

X  31.75  -  63.5  watts. 


. 

Loss  in  rings  per  phase  =  —  ^-  =  21.2  watts. 

Copper  loss  in  rotor  secondary  per  phase  =  39  +  21.2  =  60.2 
watts.  This  loss  is  PR,  per  phase,  where  /  and  R  are  certain 
"  equivalent"  values  of  current  and  resistance  of  the  secondary. 
Referred  to  the  primary  on  a  1  :  1  ratio  basis  as  is  usual  in  trans- 
former calculations,  this  loss  becomes 

60.2  =  7p2r!  =  522ri. 

60.2 
.'.  TI  =  27  '     =  0.0223  ohm,  which  is  the  desired  rotor  (secondary) 

resistance  per  phase  reduced  to  the  primary. 

Although  this  calculation  is  based  on  two  assumptions, 
namely;  negligible  exciting  current,  and  perfect  mutual  in- 
duction between  primary  and  secondary,  both  of  which  are  false, 
yet  the  error  introduced  is  so  small  as  not  to  appreciably  effect 
the  correctness  of  the  results. 

The  primary  resistance  may  now  be  calculated  in  the  usual 
way.  Length  of  mean  turn  =  twice  the  gross  length  of  stator 
+  eight  times  diameter  per  pole,  =  2  X  4.7  +  8  X  2.22  = 
27.16  in. 

Total  length  of  effective  conductor  per  phase  is 

27.16  x  *5g5?  =  108.64  ft.  .      . 

1     1  (\ 

Resistance  per  1000  ft.  of  4  No.  10  wires  in  parallel  is  -~-  —  0.29 

ohm  at  60°C. 

.'.  Resistance  per  phase,  of  stator,  is 


r0  =  X  °-29  =  °-0315  ohm- 


Primary  Leakage  Reactance.  —  This  is  determined  in  the  same 
general  manner  that  it  was  done  in  the  case  of  the  alternator. 
The  facts  that  the  air  gap  is  uniform  around  the  periphery,  that 
it  is  comparatively  very  short  and  that  the  rotor  teeth,  instead 
of  being  opposite  a  sojid  pole,  are  opposite  the  stator  teeth, 


374 


ELECTRICAL  ENGINEERING 


introduce  important  variations  into  the  calculations.  The 
leakage  flux  may  be  regarded  as  that  which  passes  through  and 
around  the  four  slots  of  any  pole  and  phase  at  right  angles  to  the 
direction  of  the  main  flux,  due  to  the  current  in  those  slots,  to- 
gether with  that  surrounding  the  end-connections  due  to  the 
current  in  them.  This  flux  may,  for  convenience,  be  divided 
into  a  number  of  parts,  as  follows: 

01  =  flux  which  crosses  the  space  occupied  by  the  conductors. 

This  space  has  an  area,   a,    (Fig.   276),   per  centimeter 
length  of  conductor. 

02  =  flux  crossing  the  area,  d,  between  the 

conductors  and  the  gap. 

03  =  flux  in  the  gap  but  not  cutting  the 

rotor  conductors.  This  is  the  so- 
called  "zig-zag"  flux,  because  it 
passes  back  and  forth  across  the  faces 
of  rotor  and  stator  teeth. 

04  =  flux  around  end-connections. 

05  =  belt  leakage  flux,  due  to  the  fact  that 

the  primary  and  secondary  coils  are 
not  always  in  positions  to  react  fully  on  each  other.  This 
flux  has  to  be  considered  only  with  reference  to  rotors  with 
definite  phase  windings.  In  such  cases  the  inductance, 
L5,  due  to  it  may  be  taken  as  approximately  5  per  cent, 
of  the  total  primary  self-induction. 
Calculation  of  inductances  due  to  0i  and  02  give 


FIG.  276. 


LI  =l47rsn: 


36 
d 


Z/2  =  4?rsn2- 
e 

cm.,  per  cm.  gross  length  of  stator,  where  s  =  slots  per  pole  and 
phase,  and  n  —  effective  conductors  per  slot. 
Similarly, 

i/* 


where  -^  is  the  effective  magnetic  conductance  of  the  path  of 

the  flux  03,  per  centimeter  gross  length  of  stator. 

This  inductance  is  really  the  sum  of  two  inductances,  L3  = 
L'z  +  L"s,  of  which  Z/3  is  made  up  of  the  interlinkages  per 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       375 

ampere  of  all  the  conductors  of  the  phase  with  the  flux,  </>'3, 
which  follows  the  path  from  A  to  B,  and  L"3  is  the  interlinkages 
of  the  conductors  in  the  two  inside  slots  with  the  flux  <£3",  which 
follows  the  path  from  C  to  D  (Fig.  277). 

Magnetic  conductance  across  a  rotor  slot  above  the  conductor 

is h  T,  where  the  letters  indicate  dimensions,   as  given  in 

Fig.  277,  in  inches.     This  magnetic  conductance  is  in  parallel 


LI 


FIG.  277. 

with  a  conductance  of  variable  magnitude  of  the  path  across  the 
gap  into  a  stator  tooth  and  back  again.  When  a  stator  tooth  is 

directly  opposite  a  rotor  slot,  this  magnetic  conductance  is  -r- 

*¥ 

where  t  is  the  width  of  a  stator  tooth  face.  When  the  two  slots 
are  directly  opposite,  the  magnetic  conductance  of  this  path  may 
be  neglected. 

In  the  first  case  (tooth  opposite  slot),  the  magnetic  conductance 
of  the  combined  path  is 


40 


In  the  second  case,  it  is 


m 


k 


The  average  of  these  two  values  is 

m      k  ,   J_ 
~n       h  +  80' 

The  flux  0'3  passes  through  this  path  twice,  as  the  figure  shows, 
and  it  also  passes  twice  across  the  gap  as  it  first  enters  and 
finally  leaves  the  rotor.  The  reluctance  of  these  latter  portions 

of  the  path  is  -—-- 


376  ELECTRICAL  ENGINEERING 

.'.  the  total  average  reluctance  is,  approximately, 

2 


= 


2  _    ,   20_    /flfo\' 

~k  ,    *  "  "  *  :  :  \fj 


and  the  required  magnetic  conductance  is 

— •-  (f-Y- 

p'a        M/o/ 

In  the  present  case,  supplying  values, 
m  =  0.03,   n  =  0.06,   fc  =  0.15,   h  =  0.26,   <  =  0.305,   g  =  0.02 

•      '  2  0.04 

0.5  +  0.577  +  1.91  T  0.305 
and 

1 


0.8  = 

Maximum  magnetic  conductance  of  the*  path  from  C  to  D 
occurs  when  the  stator  tooth  is  opposite  a  rotor  slot.  The  middle 
of  the  phase  is  here  opposite  a  rotor  tooth. 

In  this  case  the  magnetic  conductance  is 


t        m       k 
n  +h 

When  the  slots  are  opposite,  the  magnetic  conductance  is 

1 


0.944. 


, 


t     '  m       k 
n  +  h 

When  the  middle  tooth  of  the  phase  is  opposite  a  rotor  slot, 
the  magnetic  conductance  is 

~ — ; =  2.98. 


*  I  f^^  TI  A 

n       h      40 

The  approximate  average  magnetic  conductance  is  the  aver- 
age of  4.65,  0.944,  2.98  and  0.944.     This  is 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       377 

Inductance  due  to  end-connections  is,  as  has  been  stated 
in  the  study  of  the  alternator,  a  difficult  quantity  to  calculate 
with  close  approximation,  due  to  the  indefiniteness  of  the  mag- 
netic path  in  the  air.  In  the  case  of  induction  motors  with  wound 
rotors,  a  fair  value  to  assume  for  the  flux  per  ampere-conductor 
per  centimeter  length  of  the  end-connections  is 


poles 
diameter 

For  squirrel-cage  induction  motors  about  50  per  cent,  should 
be  added  to  this,  as  the  end-connections  of  the  rotor  may  be 
regarded  as  having  negligible  self-induction,  which  affords, 
consequently,  an  increased  path  for  the  stator  flux. 

Since  length  of  end-connections  per  coil  =  8  X  — »  the  flux 
per  coil  per  ampere-turn  is 


04  =  1.5  X  8  X 


J-  =  12  J- 
\  p  \  p 


120  A—  per  absolute  amp.-turn. 
p 

With  the  usual  double-layer  winding  of  the  stator  (primary)  the 

magnetomotive  force  per  coil  is  -77-* 

£ 

Therefore,  the  total  flux  per  pole  about  the  end-connections  of 
any  phase  is 


04  =  120 


snl    ID 
TttS' 


Linking  with  this  flux  are-«-  turns, 

.'.  the  inductance  (interlinkages  of  flux  turns  per  unit  current) 
per  phase  is 


L4  =  j    XpX~X  120^p  -J-    =  30s2n2 

The  total  primary  inductance  is,  then, 

d   ,    /fo\'  ,    7.5  s    ID 


378  ELECTRICAL  ENGINEERING 

Supplying  numerical  values, 
s  =  4;  s'  =  2;  n  =  4;  a  =  0.95;  b  =  0.34;  d  =  0.206;  e  =  0.244; 

(-)  '  =  1.25;  D  =  30.46  cm.;  p  =  6;  Z  =  11.94  cm.;  (&}  "  =  2.38 

\yo/  \{/o/ 

10  QQC 

L°  =  ~io^  [(0-931  +  °-843  +  L25  +  L73)4  +  2-38  x  2J 

=  0.000249  +  0.0000686  =  0.0003176  henry. 
Primary  leakage  reactance  is,  therefore, 

z0  =  27r/Lo  =  0.12  ohm 

Secondary  leakage  reactance. 

Self-induction  of  the  secondary  is  obtained  in  the  same  way. 
Thus  the  total  secondary  inductance  is 


Ll 


a    .  d 


where  all  quantities  are  known  except  —  • 


FIG.  278. 

In  Fig.  278  the  leakage  path  is  indicated  by  the  dotted  line 
for  the  relative  position  of  rotor  and  stator  teeth  in  which  the 
magnetic  conductance  is  maximum.  The  reluctance  in  this 
case  is 


0.2625 


3.81  +  0.422 


=  0.5, 


and 


go 


DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       379 

Minimum  magnetic  conductance  occurs  when  the  stator  and 
the  rotor  slots  are  directly  opposite  each  other.     Then 

1 


p3(max.)  =  2 


^  + 


-4-  —  — 

^  d  ~ 


The  value  SQ  is  the  magnetic  conductance  of  the  small  path 
across  the  faces  of  the  rotor  teeth  tips  opposite  a  stator  slot. 
It  is,  approximately, 

' 


-  5354  -  °'376'  "' 

and  the  average  value  of  the  magnetic  conductance  is 

/o       2  +  0.376 

=  --  -  -  =  1.2  approx. 
Qo  * 

The  rotor  self-induction   is  then    determined    by   supplying 
numerical  values  in  the  equation  for  LI  (see  page  370).     Thus, 


4TX  x  11.94 

Ll  =  -          -  +  +  "    -  4.09  X  10-henry, 


Reactance  =  ZirfLi  =  0.00154  ohm. 
Reduced  to  the  primary  basis,  this  gives 

(9CQ\  2 
-7^-j     =  0.0935  ohm. 

The  only  quantities  remaining  to  be  determined  are  Im  and 
Ih,  the  magnetizing  and  core-loss  currents.  The  core  loss,  or 
watts  lost  in  hysteresis  and  eddy  currents,  is  E2g,  where  Ih  =  Eg. 

This  loss  is  almost  entirely  in  the  primary  core  where  the 
normal  frequency  of  reversal  of  the  flux  obtains. 

The  secondary  core  loss  is  neglected.1 

As  with  the  alternator,  the  eddy  current  loss  will  be  taken  as 
equal  to  one-half  of  the  hysteresis  loss. 

To  find  the  latter,  then,  the  volume  of  stator  teeth  is 

Vt  =  72  X  4.23  X  0.45  =  137  cu.  in., 
where  0.45  =  area  of  a  tooth. 

1  It  is  neglected  only  in  the  case  that  the  pulsation  in  flux  due  to  the  passing 
of  the  teeth  is  negligible,  which  is,  however,  not  always  the  case. 


380 


ELECTRICAL  ENGINEERING 


Weight  of  teeth  =  137  X  0.28  =  38.4  Ib. 
Hysteresis  loss  in  teeth  at  2.8  watts  per  Ib.  at  60  cycles  and 
64,500  lines  per  sq.  in.  is 


(04  000\  L 
ffl^jj) 


170  watts. 


Outside  diameter  of  the  stator  =  16.375  in. 
Inside  diameter  of  the  stator  =  14.34  in. 
Volume  of  stator  core  is 

14.34 


Weight  of  core  =  204  X  0.28  =  57  Ib. 

Flux  density  =  516'000  -^  (1.0175  X  4.23)  =  60,000  lines  per 
f 

sq.  in. 

.'.  Hysteresis  loss  in  core  is 


57  X  2.8  X  (6°' 


1.6 


=  140  watts. 


\64,500y 

.'.  Core  loss  =  1.5(170  +  140)  =  465  watts. 

465 
Core  loss  per  phase  =  -y-  =  155  watts  =  E2g  =  EIh,  and  Ih 

155 
=  —j=r  =  2.44  amp. 

244 
Per  cent.  Ih  =  =£-  =  0.047. 

Da 

.  That  portion  of  the  magnetizing  current  required  by  the  air 
gap  has  been  assumed  as  20  per  cent.,  or  10.4  amp.  That 
required  by  the  iron  part  of  the  flux  path  must  be  worked  out 
from  the  known  densities  and  dimensions  of  the  magnetic  circuit. 
The  usual  tabulation  is  as  follows: 


Part 

Density 

A.T.  per  in. 

Length 

A.T. 

Stator  teeth 

Rotor  teeth 

Stator  core  .  . 
Rotor  core  .  . 

Total  ir< 

Smax 

84,000 
72,500 
97,000 
88,000 
60,000 
60,000 

12  °)    97 

7.4(    97 

28.  Ol 
15.  0/21-5 
4.5 
4.5 

1.15 

0.6 

4.0 
2.0 

11 

13 

18 
9 

51 

.    
mm  

max 

min  

3n  amp  -turns 

DESIGN  CONSTANTS  OF  INDUCTION  MOTOR       381 


Since  there  are  eight  turns  per  pole  per  phase,  the  magnetizing 
current  required  for  the  iron  is  calculated  from  the  equation  for 
armature  reaction. 

Arm.  reac.  =  V2  X  1.5  X  amp.-turns  per  pole  and  phase 
=  2.12  X  8  X  amp.  =  51. 

.*.  Current  required  for  the  iron  part  of  the  magnetic  path  = 
51         3. 


16.96 
Total  magnetizing  current  per  phase  is 

Im  =  10.4  -f  3  =  13.4 
per  cent.    Im  =  0.26. 

Performance  curves  of  the  motor  may  now  be  worked  out  from 
the  constants,  reduced  to  percentages,  as  follows: 

a°315X52  =  0.0258; 


=  0.0183;  x0  =  0.0982;  Xl  =  0.069;  / 


0      0.2      0.4     0.6     0.8     1.0     1.2     1.4     1.6     1.8     2.0     2.2     2.4 
Output  (Pm-f) 

FIG.  279. 


382 


ELECTRICAL  ENGINEERING 


Tabulating  as  in  the  problems  of  the  previous   chapter,  the 
resulting  values  are  as  follows: 


Slip 

0.0 

0.01 

0.02 

0.03 

0.05 

0.08 

0.1 

0.3 

0.5 

1.0 

/o 

0.257 

0.626 

1  .  107 

1.57 

2.38 

3.32 

3.775 

5.34 

5.64 

5.82 

T 

0.0 

0.498 

0.95 

1.337 

1.925 

2.37 

2.475 

1.675 

1.12 

0.6 

Pm-fo 

0.0 

0.483 

0.921 

1.288 

1.82 

2.17 

2.217 

1.163 

0.55 

0.0 

Pi 

0.0462 

0.5525 

1.022 

1.448 

2.11 

2.675 

2.867 

2.425 

1.953 

1.484 

E0Io 

0.257 

0.626 

1.107 

1.57 

2.38 

3.32 

3.775 

5.34 

5.64 

5.82 

Eff. 

0.0 

0.875 

0.90 

0.889 

0.862 

0.81 

0.772 

0.479 

0.281 

0.0 

P.F. 

0.18 

0.883 

0.922 

0.921 

0.887 

0.805 

0.76 

0.455 

0.346 

0.255 

App.  eff. 

0.0 

0.773 

0.83 

0.819 

0.765 

0.652 

0.587 

0.228 

0.0972 

0.0 

Performance  curves  are  given  in  Figs.  279  and  280. 


2.6 

2.4 

2.2 
2.0 
1.8 
1.6 

§  1.4 
H  1.2 
1.0 
0.8 
0.6 
0.4 
0.2 


t 


\ 


7 


cf, 


7 


0.2  0.4  0.6  0.8  1.0 

Speed  and  Current  -7-  4 

FIG.  280. 


1.2 


1.4 


1.6 


CHAPTER  XLIV 
ROTARY  OR  SYNCHRONOUS  CONVERTERS 

It  is  frequently  necessary  to  obtain  direct  current  when 
the  available  supply  is  alternating.  Such,  for  instance,  is  the 
case  with  many  electric  railways,  arc  lighting  systems,  etc. 
Changing  from  direct  to  alternating  current  is  also,  but  less 
frequently,  required.  The  work  of  conversion  from  direct  cur- 
rent to  alternating  current  or  vice  versa,  is  done  to  great  ad- 
vantage by  means  of  the  rotary  converter.  Looked  at  from  its 
commutator  end,  this  machine  is  a  direct-current  generator; 
from  the  other  end,  it  appears  to  be  a  synchronous  motor  with 
revolving  armature. 

Both  receiving  and  transmitting  electrical  energy,  it  is  motor 
and  generator  combined  into  one.  If  it  be  driven  by  mechanical, 
power,  it  may  give  out  electrical  power,  in  the  form  of  either 
direct  current  or  alternating  current,  or  both  at  the  same  time. 
Under  its  various  phases  of  operation,  then,  the  rotary  converter 
may  work  as  a  direct-current  generator,  direct-current  motor, 
synchronous  motor  or  alternator.  In  any  case  its  performance 
becomes  that  of  one  of  these  machines  or  a  combination  of  them, 
and  may  be  studied  in  that  light. 

Voltage  and   Current  Ratios. — The  fundamental  equation  of 

a  direct-current  generator  has  been  found  to  be  E  =  -r^g-  volts, 

where  /,  <f>  and  t,  are  respectively,  the  frequency  of  alternation 
of  voltage  in  the  conductors,  the  flux  per  pole  cutting  the  con- 
ductors and  the  total  number  of  turns  between  brushes.  If 
the  armature  is  tapped  at  two  symmetrical  points  (for  each  pair 
of  poles),  and  the  taps  are  brought  out  to  slip  rings,  then  the 
voltage  across  the  slip  rings,  by  the  fundamental  equation  for 

alternating-current  generators,  is  Eeff.  =      -.^8     effective  volts. 

This  equation  holds,  however,  only  for  concentrated  windings. 
For  distributed  windings,  it  is 

2 

E'ff.  ~  ~  > 

383 


384  ELECTRICAL  ENGINEERING 

and,  substituting  for  4.44  its  derivation,  \/2ir)  the  maximum 

,-      2       V^TT/V^      4f4>t 
voltage  is  Em  =  v  2  X  -  X      1r)8 —  =  -^—  volts,  which  is  the 

same  as  for  direct  current. 

In  calculations  involving  the  voltage  it  is  convenient  to  deal 
with  the  voltage  to  neutral,  that  is,  one-half  the  direct-current 
or  alternating-current  single-phase  voltage,  and  the  phase  volt  age 
with  polyphase  connection.  Thus,  direct-current  voltage  to 

E      2f<t>t 
neutral  is  En  =  -~  =  -TT^. 

Single-phase  effective  voltage  to  neutral  is 

En          E 

V2       2\/2 

In  a  three-phase  machine  maximum  voltage  between  rings,  = 
\/3  En  =  \/3  X  Y^-  =  \/3En. 

In  an  TV-phase  machine, 

Em  =  2En  sin  ^  =  E  sin  - 
is  maximum  voltage  between  rings. 

Effective  voltage  is  Ee/f,  =  \/2Ensiu  —  —  —j=.  sin  — 

Current  relations  are  determined  by  assuming  100  per  cent, 
efficiency,  or  input  =  output.  The  direct-current  output  is 
2EnI,  where  I  =  direct-current  line  current. 

The  input  is  N—^Ip)  for  N  phases,  where  Ip  =  phase  current. 
Equating  output  to  input,  and  solving  for  Ip) 

2NEJP  sin  -^ 
2JU -*-%,-;    -7f— 

2EJ  V%I 


sin  -.      N  sin  ~ 


Effective  line   current   is   the  vector  sum   of  two  adjacent 
effective  phase  currents.     This  may  be  proven  to  be 

2V2J 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        385 
Thus,  for  three-phase,  effective  phase  current  is 


and  effective  line  current  is 

7 


Problem  110.  —  Assuming  the  direct-current  voltage  and  current  to  be  E 
and  I  respectively,  deduce  effective  phase  voltages  and  currents  for  one-, 
two-,  three-,  four-,  and  six-phase  alternating  input. 

For  single-phase,  the  values  have  already  been  indicated.     They  are 


They  cannot  be  obtained  from  the  general  formulae  directly,  but  are 
easily  found  from  the  power  equation.  The  general  formulae  apply  to 
polyphase  machines. 

Voltage  Control.  —  Under  the  ideal  conditions  assumed,  there 
is  no  voltage  drop  in  the  armature.  Practically,  however,  there 
is  a  drop  proportional  to  the  load.  With  a  constant  impressed 
alternating-current  voltage  on  either  the  rotary  terminals  or  the 
supply  generator,  the  drop  in  the  armature,  or  in  the  armature 
and  line,  as  the  case  may  be,  causes  a  variation  in  the  terminal 
direct-current  pressure. 

It  is  frequently  desirable  to  compensate  for  this  voltage  drop 
in  line  and  armature,  in  order  to  maintain  constant  direct-current 
voltage  across  the  brushes  of  the  rotary 
converter.  One  method  of  doing  this 
is  by  proper  "  compounding  of  the  series 
field."  In  general,  the  procedure  is  to 
"  under-excite  the  shunt  field,"  so  that 
at  no  load  the  machine,  acting  as  a 
synchronous  motor,  will  draw  lagging 
current  of  about  one-third  normal 
value.  At  %-load  the  current  will  be  FIG  2gl 

in  time-phase  with  the  voltage. 

At  full-load  the  current,  due  to  the  series  field  m.m.f.,  will  lead 
somewhat,  with  the  consequent  increase  in  terminal  voltage. 

Heating  of  the  Armature.  —  In  Fig.  281  is  shown  one  phase  of  a 
rotary  converter,  whose  center  line  makes  an  angle  (0  —  a) 
with  the  field  axis.  Let  any  coil,  c,  be  displaced  an  angle,  a,  from 
the  center  line  of  this  phase,  or  0°,  from  the  field  axis. 

25 


386  ELECTRICAL  ENGINEERING 

To  Find  the  Heating  in  this  Coil. — Let  the  current  and  voltage 
be  in  time-phase  with  each  other.  Then  maximum  current  will 
flow  in  the  phase,  and  therefore  in  the  coil,  when  6  —  a  =  0. 

Thus  the  current  in  the  coil  is  not  maximum  when  the  coil  is  on 
the  axis,  but  when  the  center  of  the  phase  is  on  the  axis.  There- 
fore the  alternating  current  in  the  coil  may  be  written 

i  =  Im  cos  (6  —  a). 

If  /  is  the  direct  current  in  the  line,  ~  is  the  direct  current  in 

48 

the  coil. 

Therefore,  the  resultant  current  in  the  coil  is 

/  / 

lr  =  l  ~~  2  =        COS  '    ~~  a'  ~~  2* 

But  it  is  known  that,  in  general, 

7 

1m   = 


N  sin  ~ 


Therefore, 


The  heating  of  the  coil  is  proportional  to  the  square  of  ir',  the 
average  heating  to  the  average  square  of  the  resultant  current 

taken  over  a  half  -cycle  from  0  =  —  ^to  0  =  ^-     Thus,  average 
heating  of  coil  = 


Comparing  this  heating  to  that  of  a  direct-current  generator 
with  the  same  current  output  gives 

heating  of  rotary  coil     _  1  fi  A  cos  (0  —  a)        x  2 
heating  of  d.-c.  gen.  coil  ~  TT  '     '  -  l\  d0. 


Problem  111. — Prove  that  this  ratio  is: 


8  16  cos  a 

7T     +  TT 

Nz  sin2  jj  NTT  sin  -^ 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        387 

Proof.— 

/»-  -        _ 


/»i-    /4co3(0-_g)  _    v 
J_,\     *-»  Jl 


cos2  (0  -  a)  -  8AT  cos  (0  -  a)  sin  ~  +  N*  sin2  ^\ 

]de 


7T 

-  T2    (l6cos2(0-a)-8Arsm^cos  (0  - 


_ 

2 

Integrating  the  terms  separately, 


/*-  /•- 

/    2  /    2 

(1)  I       16  cos2  (6  —  a)d6  =  16  I       (cos2  6  cos2  a  +  2  cos  ^  sin  0  cos  a  sin  a 
^~  \  J~\ 

+  sin2  a  sin2  0)  d0 

f  /sin  26       0\  /-  cos  20\ 

=  16   cos2  a  (  —  ^  --  h  2  )  +  2  cos  a  sln  a  (  -  4  -  ) 

-sin20 


=  8r  COS2  a  +  0  +  Sir  sin2a  =8rr(sin2  a  +  cos2  a)  =8*-. 

(2)  j   2  -  8^  sin  -^  cos  (0  -  «)d0 

•'""I 

=  —  8JV  sin  -^     cos  a  sin  0  +  sin  a  (  —cos  0)    2  r 


sin  -    cos  a. 


XI  r  -,* 

N2  sin2  ~  de  =  \N*  sin2  T?  ^       ^ 
»  L  J~2 


(3)  |     N2  sin2  ^  d0  =  |  N*  sin2  -^  9  I*  ,  =  TrAT2  sin2  ^- 

J       2 

8                       16  cos  a 
/.  Ratio  =  - + 1  -  -     -• 

N2  sin2  ^  NTT  sin  -ft 

Prove  that  this  ratio  is  maximum  for  that  coil  of  the  phase  for  which  a  = 
?r,  that  is,  for  the  largest  possible  value  of  a,  and  that  the  ratio  is  minimum 
for  a  =  0. 


388  ELECTRICAL  ENGINEERING 

Problem  112.  —  Find  the  ratio  between  maximum  and  minimum  heating  on 
three-phase,  four-phase,  and  six-phase  rotary  converters. 


So^ion.-Three-phase  max.  =          .      O  +  1  -  ~  =  1.21. 


8  16  cos  0 

Three-phase  mm.  =  6QO  +  1  -  O  =  0.226. 


Three-phase  ratio,  =  =  5.35. 


8  16  cos  45° 

Four-phase  max.  =  16  sin2  45o  +  1  -  4?r  sin  450  =  0.728. 

Four-phase  min.  =  Jfl  g.*2  450  +  1  -  5^5^  =  0.2. 

max.       0.728 
Four-phase  ratio,  —r        =  --  =  3.64. 


8  16  cos  30° 

Six-phase  max.  =  36  gin2  30<>  +  1  -  67rsin30o  =  0.426. 

Six-phase  min.  =  o  +  1  -  O  =  0.199. 


.,     max.       0.426 
Six-phase  ratio,  =     -        =  2.14. 


Problem  113.  —  Find  the  ratio  of  average  heating  around  the  whole  pe- 
riphery to  the  heating  of  a  direct-current  armature. 

To  determine  the  ratio  of  average  heating  around  the  periphery  the  ex- 
pression must  be  integrated  for  the  ratio  of  heating  of  any  coil  of  a  phase  for 
all  possible  values  of  a,  and  this  must  be  divided  by  the  angular  width  of  the 
phase. 

Thus,  ratio  of  average  heating  is 

N  (>a  "  +  'I  i  _  8  16  cos  «\ 

x—  I  /  —  -  —  \da 

r\fr**%  ****%) 

*/«--•£ 

80  _  16  sin  a  "i  a 


_ 
_  0.63" 


N2  sin2  -  ^2  sin2  - 

Problem  114.  —  Find  the  ratio  of  average  heating  to  direct-current  heating 
for  three-phase,  four-phase,  and  six-phase  input. 

Three-phase:  +  1  -  "  -  0.55. 


Four-phase:      -—      -  0-63  =  0.37. 


Six-Phase:  36W  -  °'63  -  °-259- 

In  all  of  these  problems  the  input  is  taken  at  unity  power  factor. 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        389 


When  there  is  also  a  wattless  component  of  current  in  the 
winding  the  heating  is  due  to  this  component  which  is  not  com- 
pensated for,  as  well  as  to  the  power  component.  Although  the 
components  are  at  right  angles  (Fig.  282),  the  heating  due  to 
them  is  added  directly,  since  it  is  proportional 
to  their  squares,  and  the  sum  of  the  squares 
gives  the  total  current  in  the  winding. 

Another  way  to  determine  the  heating  of  a 
rotary  converter  when  the  current  has  a 
wattless  component  is  to  resolve  the  current 
into  its  components  and  add  their  instantaneous  values. 

Let  the  power  factor  of  the  alternating  current  be  cos  $.  Then, 
if  Iw  is  the  maximum  value  of  the  wattless  component  of  the 
current  in  the  winding,  and  Ie  is  the  maximum  value  of  the  power 
component, 

Iw 


tan     = 


Thus,  Iw  is  known,  or, 


Iw  =  Ie  tan  $  = 


21 


»T        •  T* 

NsmN  . 


tan  <£. 


<f>  is  positive  for  leading  current.     Iw  is  maximum  for  6  —  a 
90- -| 

.'.  iw  =  Iw  sin  (6  —  a)  for  a  coil  displaced  a°. 
Thus,  the  resultant  current  in  a  coil,  m,  is 


4  cos  (6  —  a) 

4  sin  (0  - 

a)  tan  < 

H 

N  sin  -JT-- 
2v 

4      /cos  (0  — 

.       7T 

a)  cos  0  +  sin  (0,— 

a)  sin  0 

#~7\~ 

4 

cos  <£ 

(0  -«-</>) 

i 

, 

N  sin  -JTJ.  cos  <£ 

Problem  116. — From  this  equation  determine  the  maximum  heating  of  an 
armature  coil  compared  with  that  of  a  direct-current  generator  of  the  same 
output. 
Solution. — 

rotary  current  in  any  coil  _  ir 
d.-c.  current  ~  I_ 

2 


Ratio  of 


390  ELECTRICAL  ENGINEERING 

The  ratio  of  heating 

4  cos  (0  —  a)  __  1    ,    4  sin  (0  —  q)  tan  <f>\ 


16  sin2  (0  -  q)  _  8  cos  (0-  q) 

N2  sin2  T?  N  sin  -£ 


1   C2    /16  cos2  (0  -  q) 
V-I*      AT2  sin2  ^ 


32  cos  (0  -  «)  sin  (0  -  «)  x  8  sin  (0  -  a)  x 

tan  <£ tan  <£  \d6. 


AT2  sin2  -^  N  sin  -^ 

Integrating  and  applying  limits, 

_  I  r__§£_    4.     _i_  8  TT  tan2  0  _  16  cos  q  16  sin  q  tan  <ft-i 

r  I  N2  sin2  -^  A/'2  sin2  -^  ]V  sin  -^.  JV  sin  ^ 


8  (1  +  tan2  </>)          _  16  cos  a  +  16  sin  «  tan 


But 

1  +  tan2  0  =  sec2  0  «  ~^   and  tan  0 

and 

cos  a  cos  <f>  +  sin  a  sin  <f>  =  cos  (a  — 


ATO       .Tn  TIT       •  "" 

N  *  sin  iricos2  ^  irN  sin  ^  cos  </> 

The  maximum  heating  occurs  in  that  coil  f  on  which  cos  (q  —  0)  is  mini- 
mum. 

Problem  116. — Compare  the  average  heating  around  the  whole  periphery 
with  that  of  a  direct-current  generator. 

Solution. — For  this,  it  is  necessary  simply  to  integrate  the  above  ratio 

between  the  limits  of  a  phase,  or  between  —  jj  and  +'  •»?>  and  divide  by  ^  • 


16  cos  (a  —  <f>)     x  J 
\da. 

NTT  sin       COS 


ra=^ 
...  Ratio = ^      (—  —  : + i 

27T  y AT2  sin2 -^  cos2  0 

c/«=-^ 

Substituting  the  expression 

cos  (a  —  0)  =  cos  <f>  cos  a  —  sin  ^  sin  a,  and  integrating, 

^JT 

7VY             8a                            16  (cos  <b  sin  a  —  sin  <f>  cos  a)ia"lv 
Ratio  =  =- h  « 

r  [ tf2  sin2  ^  cos2  0  NTT  sin  ^  cos  0 

~ 


ROTARY  OR  SYNCHRONOUS  CONVERTERS    391 


N      SX~N 

0          16  cos  < 

+  lir 

>  (2  sin 

jj  \  —  sin  </>  (0) 

2ir  N2  sin2  ^ 
8 

COS2  0         N 

NTT  sin 
8 

-^cos  0 
—  n  AS 

AT2  sin2  -^  cos2  0  JV2  sin2  -^.  cos2  </» 

Problem  117. — Calculate  maximum  and  minimum  heating  for  a  power 
factor  of  0.9,  lagging  and  leading,  for  three-phase,  four-phase,  and  six-phase 
rotaries,  also  the  average  heating. 

Solution. — Maximum  heating  is  obtained  by  substituting  a  =  -^  and 
minimum  heating  by  substituting  a  =  0,  in  the  equation 

8  16  cos  («  —  0) 

ratio  = H  1  ~ 


For  average  heating, 


N2  sin2  v?  cos2  <f>  irN  sin  -     cos  <f> 


ratio  =  --  0.63, 


N*  sin2    r  cos2 


where  N  =  3,  4,  6,  and  cos  <t>  =  0.9  and  —  0.9. 

The  final  substitution  of  values  is  here  left  to  the  student. 

A  good  physical  idea  of  the  distribution  of  current  in  the 
windings  of  a  rotary  armature  may  be  obtained  from  diagrams 
showing  the  direct-current  output  as  a  constant  fixed  in  space 
with  reference  to  the  brushes,  while  the  alternating-current  input 
varies  according  to  the  angular  position  of  the  phases,  by  the 
equation  i  =  Im  sin  6. 

Let  6  vary  by  steps  of  30°,  and  write  down  on  the  diagrams 
the  values  of  both  the  direct  and  alternating  current,  with  direc- 
tion arrows,  and  also  the  resultant  current  in  each  part  of  the 
winding.  This  may  be  done  both  for  single-phase  and  three- 
phase  armatures. 

If  the  unidirectional  line  current  is  100  amp.,  then  the  maxi- 
mum alternating  current  is  200  amp.  for  single-phase  and  133 
amp.  for  three-phase. 

These  diagrams  are  given  in  Fig.  283. 

Voltage  Control.  —  The  voltage  of  rotary  converters  for  railway 
work  is  controlled  by  variation  of  the  phase  of  the  current  input. 
In  general  the  machine  is  over-compounded  to  neutralize  the 
drop  in  feeders.  Therefore,  the  alternating-current  voltage 
impressed  on  the  rotary,  must  vary  to  make  up  the  necessary 
over-compounding  as  well  as  the  drop  in  the  armature. 


392 


ELECTRICAL  ENGINEERING 


The  phase  of  the  current,  as  in  a  synchronous  motor,  depends 
on  the  field  excitation.  A  reactive  coil  is  placed  in  series,  as 
shown  in  Fig.  284.  If  the  current  is  lagging  there  will  be  a  big 
drop  in  the  line  and  coil,  and  the  voltage  impressed  on  the 
rotary  will  be  reduced. 

Current    Relations    in    Single -Phase    Converter 
0°  30°  60°  90° 


100 


100 


100 


Three  -  Phase    Converter 
30  60  90 


120 


FIG.  283. 

If  the  series  field  m.m.f.  is  so  adjusted  as  to  give  leading 
current  at  full-load,  then  at  light-load,  the  current  will  be  lagging. 
By  this  means  of  adjustment,  a  constant  direct-current  voltage 
may  be  maintained.  To  get  this  adjustment,  the  shunt  and 


ROTARY  OR  SYNCHRONOUS  CONVERTERS    393 


series  field  m.m.f.  should   give   unity    power  factor  at  about 
%-load. 

Question. — In  a  given  case,  it  may  be  required  to  obtain  con- 
stant direct-current  voltage  from  a  given  constant  generator 
voltage,  EQ,  of  such  a  value  as  to  require  the  interposition  of 
transformers.  How  shall  the  transformer  ratio  be  determined, 
so  as  to  give  the  most  efficient  operation? 


Line 


Reactive  Coil 


FIG.  284. 

If  constant  voltage,  e,  impressed  on  the  rotary  is  assumed, 
the  fundamental  equation  is 

Eo  =  e+  (i  +  ji')(r  +  jx) 

where  r  -f  jx  =  impedance  of  line  and  coil. 

But  generally,  e  is  not  constant,  but  should  vary  with  the 
load — that  is,  with  i.    The  equation  is 

EQ  =  e  +  ik  +  (i  +  ji')(r  +  jx) 

=  e  +  ik  +  ir  -  i'x  +  j(ix  +  i'r)  (128) 

where  k  is  the  percentage  of  over-compounding. 

Assuming  that  i'  =  0  at,  say,  %-load,  the  first  step  is  to  find 
i'  at  no-load,  to  ascertain  that  it  is  not  excessive. 

Let  e  +  ik  =  e'.     Then  (128)  becomes 
Eo    =  e'  +(i+ji')(r+jx) 
EQ2  =  (e  +  ik  +  ir  -  i'x)2  +  (i'r  +  ix)2 

=  e2  +  I2Z*  +  i2k2  -f-  2ei'(&  -f  r)  —  2^'x  approx. 

Let  i  =  i'o  at  non-inductive  load.     Then 

E20  =  e2  +  t2(Z2  +  k2)  +  2ei0(k  +  r) 

.'.  —  =  1  +  iQ2 1 —  -h  — ?  (fc  -f  r)  (129) 


At  no-load  i  =  0. 

/.  E02  =  e2  +  ^/2^2  -  2eifx  =  e2  -  2ei'x  (approx.) 

2ex        =  ~2xL\T/     ""     j' 


394  ELECTRICAL  ENGINEERING 

T? 

The  value  of  —  is  obtained  from  Eq.  129. 

£ 

Numerical  Application.— Let  e  =  1,  r  =  0.10,  x  =  0.30,  k  = 
0.10,  i  =  1  at  full-load,  and  iQ  =  0.75. 

TjJ 

Then  —  =  1.12  and  i'  at  no-load  =  -  0.38. 

6 

The  generator  voltage  should  thus  at  no-load  be  12  per  cent, 
higher  than  the  desired  voltage  at  the  rotary  at  no-load,  or,  in 
other  words,  the  voltage  at  the  line  coming  into  the  substation 
should  be  12  per  cent,  higher  than  is  actually  wanted  at  the 
collector  rings  of  the  rotary.  The  rotary  will  reduce  the  no- 
load  voltage  by  taking  a  large  lagging  current  (in  this  case  40 
per  cent,  of  full-load  current). 

Thus  if,  for  instance,  a  single-phase  500-volt  rotary  is  to 
operate  from  a  10,000-volt  generating  station,  the  transformer 

1 0  000  28  2 

ratio  would  not  be  =  28.2,  but  =  25.2. 


If  /i  is  the  secondary  current  due  to  a  certain  phase  of  the 

primary,  whose  induced  e.m.f.  is  6;,  then  the  power  component 
of  /i  has  been  shown  to  be  e^ai. 

—  jli  is  evidently  the  secondary  current  due  to  a  primary 

phase  which  is  90°  in  space  and  time  behind  the  former.  The 
induced  e.m.f.  of  this  phase  is,  of  course,  —  je^  and  the  flux 
causing  the  e.m.f.  is  90°  in  time  ahead  of  the  e.m.f.  Thus  the 
flux  which  reacts  on  the  energy  component  of  the  original  second- 
ary current  is  proportional  to  j(—  jei)  =  ke*. 

.'.  the  torque  is  kdei 


Voltage  Control  by  Use  of  the  " Split  Pole."— The  " split  pole" 
converter  has  found  extensive  application  more  especially  in  the 
field  of  lighting  service,  where  the  fluctuations  of  the  load  are 
not  so  persistent  as  they  are  in  railway  work. 

It  is  based  upon  the  principle  that  the  voltage  ratio  is  altered 
by  shifting  the  brushes. 

Ordinarily,  the  brushes  on  a  direct-current  generator  are  set 
at  or  near  the  neutral  point,  and  when  they  are  in  that  position, 
the  generator  gives  maximum  voltage.  If  shifted  to  any  other 
point,  the  voltage  is  E  cos  a,  where  a  is  the  angle  between  the 
neutral  axis  and  that  of  the  brushes,  and  E  is  the  voltage  when 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        395 


the  brushes  are  on  the  neutral  axis.     Under  conditions  of  variable 
shift,  the  voltage  ratio  of  a  rotary  converter  is  then, 

Ea.c 


Ecos 


a 


FIG.  285. 


In  order  to  increase  the  amount  of  the  variation  of  this  ratio, 
the  poles  are  slotted,  or  split,  directly  over  the  position  chosen 
for  the  brushes,  as  shown  in  Fig.  285,  thus  enabling  the  brushes 
to  take  a  position,  which,  otherwise,  would 
be  impossible  owing  to  bad  commutation. 

Such  an  arrangement,  while  it  permits 
a  great  shift,  does  not  give  variable  shift. 
To  obtain  variation  of  the  shift,  which  is 
indeed  necessary  in  order  to  have  voltage 
control,  the  poles  may  be  split  into  two  or 
more  separate  sections,  each  with  its  own  exciting  coil. 

Suppose,  for  instance,  an  arrangement  such  as  shown  in  Fig. 
286.  Let  the  poles  N  and  S,  alone,  be  excited.  The  axis  of  the 
main  flux  is  then  along  the  line  making  an  angle  a  with  the  hori- 
zontal. Suppose,  now,  as  the  load  comes  on,  poles  Nf  and  Sf  are 
excited  by  means  of  a  series  winding.  The  flux  axis  is  then  shifted 

toward  the  horizontal,  decreasing  a, 
and  consequently  increasing  the  direct- 
current  voltage.  The  amount  of  such 
variation  of  voltage  is  considerable. 
Tests  on  a  certain  three-phase  machine 
showed  that  with  constant  impressed 
alternating-current  voltage  of  194  the 
direct-current  voltage  ranged  from  317 
to  200,  giving  at  all  times  sparkless 
commutation. 

Poor  commutation  in  rotary  converters  is  due  almost  entirely 
to  self-induction  of  the  coil  short-circuited  by  the  brushes.  This 
effect  is  of  sufficient  importance  to  justify  the  use  of  inter-poles. 

Armature  Reaction  in  Rotary  Converter 

Problem  118. — Prove  that  in  an  ordinary  rotary  converter  the  resultant 
armature  reactive  m.m.f.  due  to  the  direct  current  and  the  power  component 
of  the  alternating  current  is  zero. 

Solution. — Consider  a  two-pole  machine.    Let  /  =  direct  current,  and 

m  =  total  turns.     Then  the  ampere-turns  due  to  direct  current  =  m  X 


FIG.  286. 


396 


ELECTRICAL  ENGINEERING 


As  these  turns  are  distributed,  the  direct-current  armature  reaction  = 
2  I  =m/ 

This  reaction,  considering  the  brushes  to  be  at  0°  shift,  is  along  the  axis 
of  the  brushes. 

Armature  reaction  due  to  the  power  component  of  the  alternating  current, 
is  also  along  the  brush  axis,  but  opposite  to  the  direct-current  reaction. 
Alternating-current  armature  reaction  due  to  the  wattless  component  is 

-*  /O     N/    />"'  \S    W 

f\      V  •*    S\    >>  a.c.    /N    •*•  0 

I 

where  i'a.e.  —  Ia.c.  sin  a,  and  T0  =  effective  turns  per  phase. 

With  the  power  component,  the  expression  for  armature  reaction  is 
exactly  the  same  but  the  direction  of  the  reaction  is  along  the  brushes  in- 
stead of  along  the  field  axis  as  with  the  wattless  component. 

.'.  Alternating-current  armature  reaction  =  N  V2  ia.c.  T0  where  ia.c.  is 
the  power  component,  in  the  direction  of  the  direct  current. 

But 

V2/ 

l>a.c.   — • 


Substituting  this 


^  V2l  a.c.  To  = 


IT, 


Effective  turns  per  phase,  T0  =  kt,  where  t 

chord       N 
k  =  ratio, 


turns  per  phase. 


arc 


7slnF 


To  =       t  sin  -. 


Also, 


Nt  =  m. 

Substituting  these  values,  alternating-current 
armature  reaction  due  to  i  =  -5-  =  direct-cur- 
rent armature  reaction,  and  the  total  reaction  in 
the  brush  axis  is  0. 

Armature  reaction  is  present  in  split-pole  ro- 
taries  to  a  limited  extent.  The  direct-current 
reaction  is  always  along  the  brush  axis,  while  the 
alternating-current  reaction  due  to  the  power 
component  of  the  current  is  at  right  angles  to 
the  resultant  flux. 

The  angle,  a,  Fig.  286,  is  the  angle  of  relative  brush  shift.  The  direct- 
current  reaction,  F,  Fig.  287,  may  then  be  resolved  into  components,  F\ 
along  the  flux  axis,  and  Ft  in  line  with  the  alternating-current  reaction. 

The  alternating-current  reaction  due  to  the  power  component  of  current 
is  in  line  with  F2,  and  is  Fn  =  F  cos  a. 

.'.  F2  =  F  cos  a  =  Fn. 


Pole  Axis 


FIG.  287. 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        397 


Also, 


Fsin 


Fn 
cos  a 


sin  a  =  Fn  tan  a. 


Thus,  Fz  is  compensated  by  Fn,  while  FI  has  no  compensation  except  in 
so  far  as  this  is  accomplished  along  the  lines  of  the  main  field  excitation. 
There  remains  an  uncompensated  component  F3j  along  the  brush  axis,  which 

is 

F,  =  IP,  sin  « 


cos  a 

Example.  —  Let  .a  =  30°. 

Then  sin  a  =  sin  30°  =  0.5.     sin2  a  =  0.25.  cos  a  =  0.866. 

0.25 


Fn  X 


0.866 


0.29Fn. 


That  is,  the  uncompensated  armature  reaction  amounts  to  about  30  per 
cent,  of  the  alternating-current  reaction. 

Transformer  Connections  for  Rotary  Converters. — Most  rotary 
converters  receive  energy  from  three-phase  supply.  In  most 
cases  it  is  simply  a  matter  of  connecting  to  the 
transformer  terminals  as  would  be  done  in  the 
case  of  a  synchronous  motor.  However,  it  is 
often  economical  to  operate  the  rotaries  as  six- 
phase  machines,  receiving  the  energy,  however, 
as  three-phase.  In  connecting  six-phase,  the  principle  is  to 
always  have  the  direction  arrows  as  shown  in  Fig.  288,  for  either 
A  or  Y-connection.  This  is  illustrated  in  Fig.  289  by  what  is 
called  the  double  A-connection.  Each  primary  supplies  power 
for  two  secondaries  whose  terminals  are  led  to  the  slip  rings  of 

2,5 


FIG.  288. 


i  —  ,1*35?") 

12  11  10  9 

Sec. 

FIG.  289. 


FIG.  290. 


the  rotary  converter.  There  are  six  slip  rings.  Each  slip  ring  is 
connected  to  two  transformers.  The  terminals  are  indicated  by 
corresponding  numbers.  Fig.  290  illustrates  the  double  T-con- 
nection.  The  arrows  show  the  only  essential  precaution  that  is 
necessary  to  take.  There  are  still  other  ways  of  connecting  six- 
phase,  as  the  "diametrical,"  in  which  the  terminals  of  each 


398  ELECTRICAL  ENGINEERING 

transformer  are  connected  to  diametrically  opposite  points  on 
the  armature. 

Synchronous  Condensers.  —  When  a  rotary  converter  or  syn- 
chronous motor  is  over-excited  so  that  it  takes  a  leading  current, 
it  may  be  used  on  a  system  for  the  purpose  of  improving  the  power 
factor. 

Such  uses  are  of  frequent  occurrence  where  the  load  consists 
principally  of  induction  motors  with  their  large  lagging  compo- 
nents of  current.  Machines  used  for  the  purpose  of  neutralizing 
this  lagging  current  by  drawing  on  the  supply  for  an  equal  leading 
current,  are  called  "synchronous  condensers." 

The  commercial  problem  is,  in  any  case,  to  determine  whether 
the  saving  in  line  and  generator  loss,  improvement  in  regulation, 
and,  with  initial  installations,  the  saving  in  capital  expenditure, 
justify  the  additional  expenditure  required  for  the  installation  of 
synchronous  condensers. 

As  a  concrete  illustration,  consider  a  system  with  poor  voltage 
regulation.  Can  the  owners  afford  to  buy  synchronous  con- 
densers, at  say,  $10  per  kv.a.,  in  order  to  improve  the  operation  of 
the  system?  Let  it  be  assumed  that  the  power  factor  is  ordinarily 
only  70  per  cent.,  that  the  cost  of  energy  to  the  station  is  Ic. 
per  kw.  hr.,  and  that  the  load  factor  is  30  per  cent. 

At  normal  full-load,  i  =  1  =  load  component  of  the  current. 
Then,  since  cos  <f>  =  0.7,  if  =  1  =  wattless  component,  lagging, 
and  the  total  current  is  /  =  i  —  ji'  =  1  —  jl. 

To  neutralize  the  lagging  component,  a  leading  component,  i'c 
=  1,  is  required.  If  this  is  the  full-load  current  of  the  condenser, 
its  rating  is  at  once  determined  and,  thereby,  its  cost.  It  would 
probably  be  better  not  to  try  to  make  the  current  lead  by  90°. 
For  instance,  let  the  condenser  also  do  some  work,  say  i'c  =  0.1. 
Then 

Ic    =    0.1    +jl, 

and 


Ic  =       lM  =  1-005. 

Thus  the  condenser  current  is  practically  unaffected  in  amount 
by  the  addition  of  a  10  per  cent.  load. 

The  total  station  load  is  now  i  +  ie  =  1.1.  Whereas,  pre- 
viously its  current  was  /  =  1.4,  it  is  now  only  1.1,  and  yet  the 
useful  load  is  not  only  the  same  but  is  10  per  cent,  greater. 
The  voltage  of  the  load  is  taken  as  e  =  1. 


ROTARY  OR  SYNCHRONOUS  CONVERTERS        399 
The  line  loss  has  been  reduced  in  the  ratio  (         ,  or  38  per  cent. 


Assuming,  previously,  a  line  and  generator  loss  of  20  per  cent. 
which  would  be  reasonable  under  the  conditions,  the  saving 
amounts  to  0.38  X  0.2  =  0.076  watt. 

At  30  per  cent,  load  factor,  this  saving  is  0.076  X  0.3  =  0.0228. 
The  gain  in  power  from  the  condenser  is  0.1  X  0.3  =  0.03. 
At  Ic.  per  kw.  hr.,  and  assuming  7200  hr.  per  year,  the  value  of 
energy  saved  is 

$0.00001  X  7200  X  0.0528  =  $0.0038 
Cost  of  condenser  at  $10  per  kv.a.  is 

$0.01  X  1.0  =  $0.01 

Interest  at  6  per  cent,  on  cost  of  condenser  =  0.01  X  0.06 
=  $0.0006. 

Problem  119.  —  Apply  the  above  reasoning  to  a  system  in  which  the  normal 
load  current  is  1000  amp.  at  250  volts,  stating  the  conclusion  with  reference 
to  the  advisability  of  buying  synchronous  condensers.  Considering  the 
line  to  have  resistance  only,  show  how  the  voltage  regulation  would  be 
affected  by  the  change.  » 

There  is  still  one  feature  of  the  use  of  synchronous  condensers  that  has 
not  been  considered.  In  the  above  example,  the  load  has  been  taken  as 
varying  to  give  a  load  factor  of  30  per  cent.,  while  the  synchronous  condenser 
was  assumed  to  supply  at  all  times  a  leading  component  exactly  neutralizing 
the  lagging  component  of  the  load.  This  assumption  of  flexibility  of  the 
condenser  is  hardly  justified.  At  the  same  time  it  would  be  undesirable 
that  the  condenser  should  draw  full-load  leading  current  continuously. 
The  field  excitation  may  therefore  be  obtained  by  compound  winding  such 
that  at  full-load  i'c  =  1,  while  at  no-load  i'e  =  —  1.  The  field  of  the 
condenser  must  therefore  be  made  to  depend  on  the  entire  load  to  be 
compensated. 

By  making  the  full-load  and  no-load  wattless  components  equal  and 
opposite,  the  smallest  condenser  is  required.  This  gives,  however,  consider- 
able no-load  line  loss. 

The  use  of  synchronous  condensers  has  now  been  discussed  sufficiently  to 
enable  the  student  to  investigate  any  given  case  and  make  an  engineering 
report  on  it. 

The  question  of  where  to  install  the  condensers,  whether  at  the  load,  or 
at  the  power  station,  is  also  of  interest,  and  should  be  discussed  by  the 
student,  reasons  being  given  why  either  position  is  to  be  preferred. 


CHAPTER  XLV 
SINGLE-PHASE  ALTERNATING-CURRENT  MOTORS 

Under  this  heading  would  naturally  be  comprised  single- 
phase  induction  motors  and  the  various  types  of  commutator 
motors.  The  development  of  the  latter  class  of  machines,  to- 
gether with  certain  inherent  defects  in  the  former,  has  had  the 
effect  of  rendering  the  single-phase  induction  motor  nearly 
obsolete.  When  a  polyphase  induction  motor  is  running,  if  one 
of  the  phase  circuits  be  opened  the  motor  will  continue  to  operate 
as  a  single-phase  machine.  Its  permissible  output  will  be  greatly 
reduced  and,  in  general,  its  characteristics  will  be  changed  for  the 
worse.  The  principal  difficulty  with  the  single-phase  induction 
motor  is  its  inability  to  start.  Special  means  have  to  be  supplied, 
such  as  "splitting"  the  phase,  that  is,  dividing  the  primary  wind- 
ing into  two  circuits,  one  of  which  is  provided  with  a  condenser  or 
resistance  to  give  a  time-phase  displacement  of  one  current  rela- 
tive to  the  other.  In  some  such  way  the  motor  is  temporarily 
converted  into  a  poor  polyphase  motor  with  just  sufficient  torque 
to  enable  it  to  start  without  load.  After  coming  up  to  speed,  a 
switch  is  operated  which  causes  the  motor  to  run  thereafter 
through  power  supplied  to  one  phase. 

The  Series  Motor. — It  has  already  been  pointed  out  that  a 
direct-current  series  motor  possesses,  fundamentally,  the  quali- 
fications for  operation  on  alternating 
current.     Practically,  in  direct-current 
motors,  the  field  is  made  strong  and 
the  armature  weak.      In    alternating- 
current  series  motors  the  reverse  is  the 
case;  the  armature  m.m.f.  is  three  or 
FIG.  291.  four  times  as  strong  as  that  of  the  field 

circuit. 

To  determine  the  values  of  flux,  current,  torque,  power,  etc., 
we  may  proceed  as  follows: 

Consider  an  armature  coil  displaced  6°  from  the  horizontal 
(Fig.  291).     The  flux  enclosed  by  the  coil  is 
</>  =  <E>m  sin  co£  sin  0 
400 


ALTERNATING-CURRENT  MOTORS  401 

since  the  flux  is  alternating,  the  total  flux  at  any  instant  being 
<J>ra  sin  co£. 
Therefore  the  induced  e.m.f.  per  coil  is 

d<f>  /  d6\ 

de  =  — rr  =  —  3?m  ( w  cos  w£  sin  6  +  sm  w£  cos  0  -77 )  • 
at  \  ail 

Let,  now,  B  =  co^,  where  o>i  =  Zirfi  and/i  =  frequency  of  rota- 
tion, i.e.,  due  to  rotation  of  the  armature;  the  coil  has  moved 
through  an  angle  B  in  the  time  t,  wi  being  the  angular  velocity  of 
the  rotation. 
Then 

dB  _ 

dt  ""  Wl' 
and 

de  —  —  3>m(a>  cos  ut  sin  B  +  wi  sin  co£  cos  B). 

T 
If  there  are  T  turns  between  brushes,  there  are  —  dB  turns  in 

the  little  element  dB.     Therefore  the  total  induced  e.m.f.  is 

T  CQ=  2^ 

e  SB I  <i>,n(a>  cos  co£  sin  B  +  coi  sin  co£  cos  0) 

*Jt— £ 


sn  co^ 

volts- 


Thus,  the  induced  e.m.f.  is  seen  to  be  of  fundamental  frequency 
I,  and  in  time-phase  with  the  flux. 

Let  the  components  of  the  flux  be  <£/,  the  field  flux  along  the 
pole  axis,  and  <t>a,  the  armature  flux  along  the  brush  axis.  These 
component  fluxes  are  then  in  space  quadrature.  Let,  also,  N/ 

be  the  number  of  field  turns  per  pole,  and  Na  the  equivalent 

2 
number  of  armature  turns  (Na  =  -T  for  distributed  winding). 

If  the  magnetic  reluctance  were  uniform  about  the  armature 
periphery,  -f  would  be  equal  to  -r/- 

<Po  f*m 

In  practice  this  is  not  the  case.  These  motors  are  provided 
with  definite  poles,  and  therefore 

^  =  m  — ^ 
where  m  >  1. 

26 


402  ELECTRICAL  ENGINEERING 

Let  Nf  =  n 

•iVa 

where  n  <  1. 

Then  $/ 

—  =  mn. 

Since,  now,  the  conductors  are  rotating  in  a  magnetic  field, 
there  is  induced  in  them  an  e.m.f.,  Er,  of  rotation,  whose  effective 
value  is 


Na,  here,  is  used  for  the  effective  turns  of  an  equivalent  single- 
phase  alternator. 

The  induced  e.m.f.  in  the  armature  due  to  the  alternation  of 
the  armature  flux,  <£a,  is 

VOltS, 


and,  similarly,  the  induced  e.m.f.  in  the  field  due  to  the  alterna- 
tion of  the  field  flux,  <£/,  is 


. 

Ef  =      ~W      "  V°ltS* 

The  total  e.m.f.  induced  either  by  rotation  or  by  "  transformer 
action"  is  the  sum  of  these  three  e.m.fs.  treated  as  time  vectors. 

Er  is  in  time-phase  with  $/,  while  Ea  and  E/  are  in  time-quadra- 
ture both  with  <f>a  and  $/. 

.'.  the  total  induced  voltage  is 


Et  = 

v  Erz  -\-  (Ea  -\-  Ef)  2. 

But, 

Er 

flNa           S 

Ef 

fNf        n 

where 

s  = 

j,  and,  as  before,  n  =  j/* 

Also, 

Ea 
Ef~ 

Na$a              1 

Nf$f  ~  mn2' 

.*.  Substituting 

•n 

Er  = 

-Ef  and  Ea  =  &y 

n    '                     mn* 

Ef    I         (ran2  +  I)2      Ef     - 


ALTERNATING-CURRENT  MOTORS  403 

If  resistance  drops  in  the  field  and  armature  coils  are  neglected, 
Ei  =  E,  the  impressed  e.m.f.     The  reactance  may  be  determined 

as  in  the  case  of  the  induction  motor.     Let  x/  =  reactance  of 

pi 

field  winding.     Then  xj  —  -/• 

Substituting  in  the  equation  for  induced  e.m.f., 


En 
I  =  -  —7=-  =  -  --=  approx. 


The  electrical  power  input  is 

Pi  =  El  cos  a. 
Mechanical  power  output  is 

P-  FT  -      En 
r  —  EJTL   — 


-p 

Torque  =  T  —  —  =  •  —  j  synchronous  watts. 

S 


torque  at  synchronous  speed 
torque  at  standstill 

T.       Ao  (mn2  +  I)2 

=  TQ  ~~  A8  ~  mV  +  (mn2  +  I)2' 

As  an  example,  assume  a  motor  of  uniform  reluctance  (m  —  1), 
and  of  the  same  number  of  turns  in  the  armature  and  field  coils 
(n  =  1). 

Then 

T  22  4 

TO  =  r+~2"2  =  5  =  °-8- 

In  such  a  machine  the  starting  torque  is  not  much  greater  than 
that  of  full-load,  which  is  not  very  satisfactory  for  the  class  of 
work  series  motors  are  usually  called  on  to  perform. 

An  approximate  ratio, 


-f 
Using  this  expression,  if  n  =  0.5, 


The  starting  torque  is  five  times  as  strong  as  that  at  synchron- 
ous speed. 


404 


ELECTRICAL  ENGINEERING 


Compensated  Series  Motor.  —  The  practical  operation  of  the 
series  motor  is  attended  with  difficulty  owing  to  the  tendency  to 

excessive  sparking.  It  also  has  poor 
power  factor.  In  the  short-circuited 
coils  heavy  electromotive  forces  are 
produced  by  the  alternating  flux.  To 
overcome  these  electromotive  forces 
compensating  windings  are  supplied 
which  under  speed  conditions  act  like 
inter-poles  to  neutralize  the  armature 
reaction  and  self-induction  of  the  short- 

circuited  coils.     Fig.  292  gives  the  wiring  diagram  of  the  com- 
pensated series  motor. 

Problem  120.  —  Derive  the  complex  equations  for  the  series  alternating- 
current  motor  with  compensating  coil. 

Solution.  —  The  impressed  e.m.f.  may  be  written 

E  =  IR  +Er  +_j(Ix  +  Ef\ 

where  R  =  total  resistance  of  armature,  field  and  compensating  coil, 
x    —  self  -inductive  reactance  of  armature  and  coil, 
Er  =  e.m.f.  due  to  rotation, 
Ef  =  induced  e.m.f.  in  the  field. 
It  has  been  shown  that 


also 

Substituting, 


1JT 

Er 


E 


\/2  0/,  and  <£/  =  — 


Nr  I 


1Q8 

where        Nf  =  field  turns  and  p  =  reluctance,  E/  = 
Also, 


effective  value  of  the  field  flux, 
4 


r^ 


reactance  of  field,  =  2ir  f  L/, 


and 


and 


_ 

I  X  H*  "  p  X  108 


Ef  =  x/7,  by  substitution, 
/,  E  =  IR  +  Ixf 


ALTERNATING-CURRENT  MOTORS  405 

or, 

E  =  /(a  -f  jo),  where  a  =  R  + -x/ 

and 

b  =  or  +  xf. 


a  +jb 
and 

i.-- £ 


Power  factor  =  cos 


A/a2  +  62 
x,I*. 
power       x/I* 


Power  =  #r/  =  -Xflz. 


Torque 


The  general  equations  for  series  alternating-current  motors  may  thus  be 
written.     They  are : 


E  =  I(R  +          2  +  (x 

% 

giving  the  voltage  and  current  relation, 
Power  factor  =  cos  a,  where 


X+  Xf 

tan  a  = 


Power  =  Erl  =  ~  X 


f     2   +   (X   +  Xf}' 


176  X  syn.  watts  X  poles 
Torque  =    -  -  ft.-lb., 

s 

power 
=          —  synchronous  watts. 

Problem  121. — A  series  alternating-current  motor  has  the  following  con- 
stants: 2  hp.,  4  poles,  60  cycles,  armature  has  29  coils,  each  coil  having  9 

2       29  X  9 
turns.     .'.  Na  =  -  X  — r —  =  41.4  effective  turns  per  pole.     Each  field 

7T  ~t 

field  turns       Nf         19 
pole   has   19  turns.     . .   Nf  =  19.     n  =  ratio  arm  turns  =  ^  =  4L6  = 

0.46.     x/  =  1.5;  x  =  0.3;  x/  +  x  =  1.8;  r«  =  0.3;  rf  =  0.14;  rcomp.  =  0.3; 
fttotaz  =  0.74;  voltage  impressed  =  110. 

Find  power  output,  power  factor,  efficiency,  apparent  efficiency,  torque, 
at  speeds  in  per  cent,  of  synchronous  speed. 


406 


ELECTRICAL  ENGINEERING 


s  —  per     cent. 

speed 

0.0 

0.2 

0.4 

0.6 

0.8 

1.0 

1.5 

2.0 

s 
n 

0.0 

0.435 

0.87 

1.3 

1.74 

2.17 

3.26 

4.35 

s 

0.0 

0.635 

1.31 

1.95 

2.62 

3.26 

4.89 

6.53 

a  =  R  +  -xf 

0.74 

1.393 

2.05 

2.69 

3.36 

4.0 

5.67 

7.27 

n 

b  =  x  +  xt 

1.8 

1.8 

1.8 

1.8 

1.8 

1.8 

1.8 

1.8 

a2  +  b2 

1.94 

er/2    o 

2.27 

A  O      C 

2.74 

A  f\     O 

3.24 

O  A      f\ 

3.8 

or*    f\ 

4.38 

OP     O 

5.92 

7.5 

Hm 

Va2  +  b2 

56.8 

48.5 

40.2 

34.0 

29.0 

25.2 

18.6 

.7 

I2 

3230.0 

2350.0 

1620.0 

1160.0 

840.0 

635.0 

346.0 

217.0 

p   _  /2   ij. 

0.0 

1535.0 

2120.0 

2260.0 

2200.0 

2070.0 

1690.0 

1420.0 

n 

PT?                " 

OOQ1 

OA1  R 

One 

OOO 

Ooojr 

OO1  A 

Of\K 

OQ7 

i   .A1.  —          /  — 

Pinput  =  I2a 

.ool 

2390.0 

.  O.LO 

3280.0 

.  /O 

3330.0 

.OO 

3130.0 

.  OOO 

2820.0 

.  »7  J.T: 

2540.0 

.  yo 
1950.0 

.  y  < 
1580.0 

Erf.  t*pr 

0.0 

0.47 

0.637 

0.723 

0.78 

0.817 

0.868 

0.9 

El 

6250.0 

5340.0 

4420.0 

3740.0 

3200.0 

2780.0 

2050.0 

1620.0 

App.  eff. 

0.0 

0.289 

0.477 

0.6 

0.69 

0.747 

0.823 

0.872 

Hp. 

0.0 

2.05 

2.84 

3.03 

2.95 

2.78 

2.27 

1.9 

Torque 

7700.0 

5300.0 

3770.0 

2750.0 

2070.0 

1130.0 

710.0 

T.  in.  ft.-lb. 

90.0 

62.0 

44.1 

32.2 

24.2 

13.3 

8.3 

0.4     0.6     0.8     1.0     1.2     1.4     1.6     W    2.0 
%  Speed,  S. 

FIG.  293. 


ALTERNATING-CURRENT  MOTORS 


407 


FIG.  294. 


The  performance  curves  of  this  motor  are  given  in  Fig.  293. 
Repulsion  Motor. — A  different  motor,  but  one  whose  charac- 
teristics are  the  same  as  those  of  the  series  type,  is  that  which 
is  known  as  the  repulsion  motor.     Its  principle  was  discovered 
by  ELIHU  THOMSON. 

Its  chief  features  are  that  the  armature  is  short-circuited 
through  the  brushes,  and  that  torque  is  obtained  by  giving  the 
latter  a  shift  of  about  75°  from  the  normal  neutral  position. 
The  angle  0,  Fig.  294,  is  that  between 
the  field  axis  and  the  brush  axis. 

If  the  brushes  were  set  on  the  field 
axis  (6  =  0°),  the  machine  would  act  as 
a  short-circuited  transformer.  If  the 
brushes  were  set  on  the  neutral  axis 
(6  =  90°),  no  current  would  flow  in  the 
armature  at  standstill  since  there  would 
be  no  resultant  induced  e.m.f.  acting  in 
the  short-circuit. 

Two  sets  of  brushes  are  sometimes  used,  one  set  being  on  the 
flux  axis  and  the  other  on  the  neutral  axis.  A  prominent  ex- 
ample of  this  type  is  known  as  the  WINTER-EICHBERG  motor. 

Considering  now  the  theory  of  repulsion  motor  action,  let  the 
flux  <f>  be  resolved  into  two  components,  fa  =  <j>  cos  0  in  the 
direction  of  the  brush  axis  and  <£/  =  <£  sin  0  at  90°  from  this  axis. 

<fo  may  be  called  the  transformer  flux 
owing  to  its  purely  inductive  action 
on  the  armature.  Let  the  magneto- 
motive forces  causing  <£,  0f  and  <£/ 
be,  respectively,./'7,  Ft  and  Ff,  and 
that  of  the  armature  be  Fa.  These 
may  be  represented  as  acting  in  a 
motor  as  shown  in  Fig.  295.  If  there 
were  perfect  mutual  induction  between 
Ft  and  Fa,  these  would  be  equal  at 
standstill,  and  the  flux  <f>t  would  cease 

to  exist.  In  that  case  there  would  be  no  xl  drop  across  the 
"  transformer "  poles,  and  if  the  rl  drop  were  neglected,  the 
whole  impressed  voltage  would  be  across  the  "field"  poles. 

Denoting  by  E  the  impressed  e.m.f.,  by  Et  the  drop  across  the 
transformer  poles,  and  by  Ef  the  drop  across  the  field  poles,  then, 
assuming  perfect  mutual  induction  at  standstill, 
Et  =  0,  and  Ef  =  E. 


FIG.  295. 


408  ELECTRICAL  ENGINEERING 

As  the  armature  starts  to  revolve  due  to  the  torque  between 
the  field  flux  and  current,  an  e.m.f.  is  generated  in  it  by  rotation. 
This  e.m.f.,  Er,  is  in  time-phase  with  the  flux  $/,  and  generates 
current,  /r,  in  the  short-circuited  armature. 

/r  must  attain  a  value  such  that   L-r  —  Er.     L-r  may  be 

called  the  e.m.f.  of  alternation  of  the  current. 

It  has  been  assumed  that  at  standstill  <fo  =  0.  With  the  pro- 
duction of  Ir  by  rotation,  $t  is  increased,  and  at  synchronous  speed 
the  total  current  reactions  cause  <f>t  to  equal  <£/. 

_^ _         These  relations  may  be  seen  with  the 

ET  0/  aid  of  a  vector  diagram,  Fig.  296.     <£/ 
,.,  and  fa  are  in  space-quadrature,  and  also, 

due  to  the  compensating  action  of  the 
FIG".  296.  armature  on  the  transformer  poles,  they 

are  in  time-quadrature  as  well.     Their 
relative  values  are  expressed  by  the  equation 


where /i  =  frequency  of  rotation  and  /  =  synchronous  frequency. 
The  instantaneous  value  of  <£/  is 

<t>f  (inst.)  =  $/  cos  6, 
and  of  fa,  is 

<j>t  (inst.)  =««!>/  sin  6. 

turns  on  transformer  poles       Nt 


Let  n  — 


turns  on  field  poles  N/ 


It  must  be  remembered  that  this  machine  is  hypothetical; 
there  are  not  actually  two  sets  of  poles.  The  relative  values  of 
Nt  and  N/  depend  entirely  on  the  brush  shift.  At  45°  shift  they 
are  equal,  and  n  =  1.  At  75°,  or  a  =  15°,  n  >  1. 

The  e.m.f.  of  rotation,  then,  evidently  depends  on  the  speed, 
the  brush  shift,  and  Ef)  the  voltage  across  the  field  poles.  It  is 

Er  =  snEf. 

These  two  e.m.fs.,  that  of  the  field  and  that  due  to  rotation  of 
the  armature,  make  up  the  total  e.m.f.,  E.  Since  they  are  in 
quadrature, 

E  =  \/Er2  +  Ef2  =  Ef\/n2s2  +  1. 


ALTERNATING-CURRENT  MOTORS  409 

The  current,  since  rl  drop  is  neglected,  is 

I  =  &. 

Xf 

On  the  basis  of  perfect  mutual  induction,  also,  the  transformer 
induced  armature  ampere-turns  =  ampere-turns  on  the  trans- 
former poles,  or  ItNa  =  INt. 


where  It  is  the  current  in  the  armature  due  to  the  inductive  action 
of  the  transformer  poles. 

Also,  the  current  in  the  armature  due  to  Er  gives 

IrNa    =    J  INf, 

that  is,  the  armature  ampere-turns  due  to  rotation  are  propor- 
tional to  the  ampere-turns  on  the  field  poles  in  the  ratio  -j*     These 

two  currents,  It  and  /r,  are  evidently  in  time-quadrature  although 
in  space-phase.     The  total  armature  current  is,  therefore, 

/.  =  /.  ITU 


Power  factor  of  the  motor  is  determined  from  the  components 
of  the  impressed  e.m.f.,  E.  Thus, 

r  E' 

power  factor  =  cos  a  =  -~-  • 

Power  developed  is  P  =  Erl. 

p 

Torque  =  r-  — j- 

synchronous  speed 

Current  input, 

"fjl  T7f 

f[j  f  f*j 

xf       xf\/n2s2  -f  1 

Where  imperfect  mutual  induction  exists  it  is  necessary  to 
introduce  the  term  xt  to  account  for  the  self-induction  of  the 
transformer  poles, 


INDEX 


Abampere,  1,  21 

Abohxn,  1 

Abvolt,  1 

Admittance,  110 

Air  gap  of  alternators,  293 

of  induction  motors,  365 
Alternating  current,  38 
Alternator,  246 

definite  pole,  258 

design,  289 

field  winding,  298 

heating,  299 

rating,  250 

reactance,  270 

regulation,  300 

round  rotor,  259 

short-circuit,  301 

three-phase,  246 

two-phase,  246 
Ampere,  1,  21 
Apparent  efficiency  of  transmission, 

117 
Armature,  drum,  48 

laminations,  296 

length,  58,  294 

reaction,  47,  60,  252,  278,  322, 
395 

resistance,  64,  295 

ring,  48 

winding,  46 
Auto-transformer,  240 

for    two-p  has  e — three-p  h  a  s  e 
transformation,  241 

with  induction  motor,  364 
Average  value  of  sine  wave,  41 

Balancer,  90 

Ballistic  galvanometer,  102 

Bar  windings,  inductance  of,  275 

Battery  for  three-wire  system,  90 

Belt  leakage  flux,  374 

Booster,  94 

British  thermal  unit,  7 

BROOKS  and  TURNER,  36 


Brush  design,  66 
resistance,  65 

Calorie,  7 
Capacity,  123 

distributed,  162 

of  a  concentric  cylinder,  155 
•      of  a  sphere,  124,  153 

of  a  spherical  concentric  con- 
denser, 154 

of  a  three-phase  cable,  158 

of  a  transmission  line,  156 

of  two  parallel  plates,  156 

reactance,  126 
Characteristic,  motor  speed,  99 

torque,  100 

Charge  of  a  condenser,  123 
Charging  current  of  a  transmission 

line,  158 

Circuit,  magnetic,  57 
Coercive  force,  19 
Commutating  pole,  74,  82 

winding,  80 
Commutation,  82 
Commutator,  66 
Compensated    A.-C.    series   motor, 

404 

Compensating  winding  74,  80,  82. 
Compensators;     see     Auto- trans- 
formers. 
Complex  quantities,  addition  of,  169 

differentiation  of,  172 

division  of,  170 

exponential    representation  of, 
172 

involution  and  evolution  of,  170 

logarithm  of,  172 

multiplication  of,  170 

representation  of,  169 

roots  of,  171 
Compounding  curves  of  alternators, 

266 
Condenser,  123 

synchronous,  396 


411 


412 


INDEX 


Conductance,  2,  110 
Conductivity,  2 

Constant-current  transformer,   121 
potential — constant-current 

transformation,  120,  140 
Core-loss  current  of  induction  motor, 

379 

current  of  transformer,  176 
dependence     on     e.m.f.     wave 

shape,  194 

relation  to  form  factor,  195 
Cosine  series,  171 
Cost  of  transformers,  223 
Coulomb,  1,  15 
Cross-magnetization,  49 
Current,  1 

density,  56 

/      in  alternator  armature,  291 
distribution   in   rotary  con- 
verters, 391 

ratios  in  rotary  converters,  383 
Cylindrical  poles,  15 

"Dead  points"  in  induction  motor, 

369 

Demagnetization,  49 
Density,  current,  56 

energy,  17 

of  magnetic  field,  14 
Design  of  alternators,  289 

of  D.-C.  generators,  55,  74 

of  induction  motors,  365 

lifting  magnets,  26 
Dielectric  strength,  125 
Distorted  waves,  133,  196,  232 
Distributed  capacity,  162 

inductance,  162 

winding,  50 

three-phase,  281 

Distribution   factor   of   three-phase 
winding,  292 

DOBROWOLSKY,  91 

Drum  winding,  48 
Ducts,  ventilating,  58,  294 

Eddy  current  loss,  187 

of  D.-C.  generator,  70 
in  transformers,  211 

Edison  three-wire  system,  89 


Effectiveness  of  coil,  42 

Effective  value,  of  distorted  wave, 

134 

of  sine  wave,  40 

Efficiency  of  D.-C.  generator,  68 
of  transformer,  210,  220 
of  transmission,  117 
Electro-dynamometer,  146 
E.m.f.  waves,  generation  of,  30 
End  connections,  inductance  of,  271, 

377 

stresses  on,  319 
Energy  density,  17 

of  short-circuit,  314 
Equalizer,  89 
Exciting  current  of  transformer,  174, 

191 
Exponential  series,  172 

Farad,  157 

FARADAY,  14,  29,  44,  124 

Field  current  at  short-circuit,  323 

voltage  at  short-circuit,  323 

winding  of  alternators,  298 
Flat  poles,  15 
Flux  calculation,  56 

for  alternators,  292 

density  in  teeth,  294,  367 
in  transformers,  212 

distribution   around    armature, 
67 

leakage,  56 

Fly-wheel  effect  on  hunting,  287 
Force,  lines  of,  14,  124 

on  wire  in  a  field,  21 

tubes  of,  124 
Form  factor,  42 

relation  of  core  loss  to,  195 
FOURIER'S  series,  190,  196 
FRANKLIN,  124 
Frequency,  38 

Friction  loss  in  D.-C.  generators,  70 
Fringing  factor,  58 
FROELICH'S  equation,  51 

Galvanometer,  ballistic,  102 

as  ammeter,  103 
GAUSS,  15 
GAUSS'  theorem,  15 


INDEX 


413 


Generator,  alternating-current,  246 

design  of  a  D.-C.,  55 

direct-current,  45 

efficiency  of  a  D.-C.,  68 

homopolar,  44 

induction,  258 

losses  in  D.-C.,  68 

three-wire,  90 

turbo-,  258 
Generators  in  parallel,  88 

in  series,  89 
Gradient,  potential,  153 

Heating  of  alternators,  299 

of  D.-C.  generators,  72 

of  rotary  converters,  385 

of  transformers,  222 
Henry,  the,  33 
Horoopolar  generator,  44 
Horsepower,  100 

of  induction  motor,  348 
Hunting,  283 
Hysteresis,  18 

constants,  187 

loop,  186,  190 

loss,  186 

in  D.-C.  generator,  70 
in  transformers,  211 

Impedance,  105,  108 
condensive,  126 
triangle,  106 
Inductance,  33 

distributed,  162 
of  bar  windings,  275 
of  concentric  cable,  159 
of  end  connections,  271,  377 
of  transformers,  202 
of  transmission  line,  119,  160 
maximum,  of  coil,  36 
Induction  motor,  342 

abnormal  operation  of,  361 

air  gap,  365  « 

at  end  of  transmission  line,  364 

core  loss,  380 
current,  379 

"dead  points,"  369 

design,  365 

equivalent  circuit,  346 


Induction  motor,  horsepower  of,  348 
magnetizing  current,  379 
maximum  output,  349 
performance  curves,  349,  381 
power  factor  of,  348 
reactance,  373 
secondary  resistance,  352 
rotary  field,  343 
single-phase,  400 
slip  of,  345 

slot  and  tooth  dimensions,  366 
squirrel-cage  winding,  362 
theory  of  operation,  344 
torque  of,  347 
types  of,  362 

with  auto-transformer,  364 
mutual,  117 

Inductive  circuit,  characteristic  fea- 
tures of,  39 
current  in,  35 
fundamental  equation  of,  38 

Instruments,  formulae  of  magnetic, 
24 

Insulation  in  slots,  291 

thickness  in  transformers,  215 

Intensity  of  electric  field,  152 
of  magnetic  field,  14 

Intel-linkage  factor,  159 

Joule,  the,  6 

KENNELLY,  113 
KIRCHOPF'S  laws,  4 

Laminations,  296 
Leakage  factor,  77 

flux,  56 

Lifting  magnet,  26 
Lines  of  force,  14,  124 
Losses  in  D.-C.  generator,  68 

Magnet,  lifting,  26 

for  metors,  16 

pull  of,  16 
Magnetic  circuit,  57 

dimensions,  296 

cycle,  17 

density,  14 

field  intensity,  14 


414 


INDEX 


Magnetic  intensity    determinations 

of,  21    , 
Magnetism,  14 

molecular  theory  of,  17 

residual,  17 
Magnetization,  18 

curves,  26 
Magnetizing    current    of    induction 

motor,  379 

of  transformers,  176,  213,  220 
Magnetomotive  force,  18 
Maximum     output     of     induction 
motor,  349 

short-circuit  current,  305 
Molecular  theory  of  magnetism,  17 
Motor,  A.-C.  single-phase,  400 

D.-C.,  98 

induction,  342 

principle  of,  20 

repulsion,  407 

speed  characteristics,  99 

synchronous,  324 

WlNTER-ElCHBERG,  407 

Multiphase  short-circuits,  320 
Mutual  induction,  coefficient  of,  177 

Natural  period  of  a  machine,  284 
Network,  solution  of,  4 

OERSTED,  20 
Ohm,  1 
OHM'S  law,  3 
ONNES,  301 

Open  delta  transformer  connection, 
236 

Parallel  circuit  calculations.  129 
Permeability,  14,  19 
Phase  characteristics   of  synchron- 
ous motors,  335 

difference,  106 
Pole  intensity,  17 
Potential,  162 

difference,  1 

gradient,  153 
Power,  6,  42 

average  value  of,  114 

by  symbolic  method,  113 

equation,  35 


Power  factor,  116 

of  A.-C.  generator,  117 

of  D.-C.  motors,  99 

of  short-circuit,  305,  318,  321 

of  three-phase  alternator,   249 
Pull  of  magnet,  16 

Rating  of  alternators,  250 

of  auto-transformers,  240 

of    T-connected    transformers, 

238 
Reactance,  39 

of  alternators,  270 
of  armature  coils,  258 
of  induction  motors,  373 
of  synchronous  motor,  324 
of  transformers,  221 
Regulation  of  alternators,  300 
of  transformers,  221     - 
of  transmission  line,  117 
Reluctance,  68 

of  three-phase  transformers,  234 
Repulsion  motor,  407 
Residual  magnetism,  17 
Resistance,  1 

armature,  64 

brush,  65 

in  induction  motor  secondary, 

352 

of  rotor,  371 

of  synchronous  motor,  324 
series  field,  65 
shunt  field,  62 
Resistances  in  parallel,  3 

in  series,  3 
Resistivity,  2 

of  conductors,  3 
Resonance,  128 
effects,  137 

Resultant  field  of  three-phase  sys- 
tem, 254 
Ring  winding,  48 
Rotary  converter,  383 

armature  reaction,  395 
current  distribution  in,  391 
heating  of,  385 
six-phase,  385,  397 
transformer  connections  for, 
397 


INDEX 


415 


Rotary    converter,  voltage  control, 
385,  391 

voltage  and  current  ratios,  383 

with  split  poles,  394 

field,  343 
Rotor  resistance  calculation,  371 

slots,  number  of,  369 

Saturation  curves,  26 

calculation  of,  59,  78 
SCOTT  connection,  237 
Series  circuit  calculation,  126 

field  winding,  63 

lighting  circuits,  120 

motor,  A.-C.,  400 

starting  torque  of,  403 

compensated  A.-C.,  404 

Short-circuit  current,  maximum,  305 

energy  of,  314 

field  current  of,  323 

field  e.m.f.  of,  323 

multiphase,  320 

of  Uternators,  301 

power  of,  305,  318,  321 

stress  on  shaft  due  to,  313 
Shunt  field  winding,  62 
SIEMENS'S   e  1  e  c  t  r  o-dynamometer, 

146 
Sine  series,  171 

wave,  average  value  of,  39 

effective  value  of,  41 
Single-phase  A.-C.  motors,  400 
Six-phase  rotary  converter,  385,  397 
Slip  of  induction  motors,  345 
Slot  design,  276 

dimensions  of  alternators,  290 
of  induction  motors,  366,  369 

insulation,  291 

pitch,  290 

Slots,  number  of  rotor,  369 
Speed  characteristics  of  motors,  99 
Split  pole  rotary  converters,  394 
Squirrel-cage  winding,  362 
Starting  torque  of  series  A.-C.  motor, 

403 

Stator  teeth  flux  density,  367 
STEINMETZ,  113,  162,  178,  187,  198 
Stresses  in  transformers,  202 
calculation  of,  206 


Stresses  on    armature  end  connec- 
tions, 319 
Susceptance,  110 
Symbolic  method,  113 
Synchronous  condensers,  398 

converters;    see    Rotary    con- 
verters. 

impedance,  302 
motor,  324 
diagram,  325 
equations,  327 
phase  characteristics,  335 
reactance,  324 
resistance,  324 
V-curves,  326,  341 
watts,  348 

Teeth  flux  density,  294,  367 
Temperature  coefficient,  2 

of  conductors,  3 
THOMSON,  ELIHU,  407 
Three-phase  system,  227 

resultant  field  of,  254 
wave  distortion  in,  230 
with  neutral,  228 
transformers,  229,  233 

reluctance  of,  234 
Three-wire  generator,  90 

system,  89 

Torque  characteristics,  100 
of  D.-C.  motors,  99 
of  induction  motors,  347 
"Transformer    action"    in    single- 
phase  motors,  402 
Transformer,  auto-,  240 

calculation,  example  of,  179 

approximate  method  of,  181 
coil  connections,  216 
connections    for     rotary     con- 
verters, 397 
core  area,  214 
length,  214 
loss  current,  176 
diagram,  175 
design,  209 
efficiency,  210,  220 
equivalent  circuit,  178 
exciting  current,  174 
flux  calculation,  214 


416 


INDEX 


Transformer,  flux  density,  212 

heating,  222 

inductance,  202 

insulation  thickness,  215 

losses,  211 

magnetizing  current,  176,  213, 
220 

number  of  turns  of,  214 

rating,  179 

reactance,  221 

regulation,  221 

weight,  223 

winding  calculations,  215 
Transformers  connected  open  delta, 

236 
three-phase,  244 

core  type,  209 

cost  of,  223 

cruciform  type,  209 

for  six  phases,  245 

in  parallel,  243 

in  series,  242 

lighting,  211 

methods  of  cooling,  210 

power,  211 

rating  of  T-connected,  238 

shell  type,  209 

stresses  in,  202 
calculation  of,  206 

T-connected,  237 
Transmission  line  calculation,   113, 

165 
approximate,  132 


Transmission  line  capacity,  156 
Tubes  of  force,  124 
Two-phase — three-phase  transforma- 
tion, 238 

Unit  pole,  14 
Units,  1 

V-curves    of    synchronous    motor, 

326,  341 
Vector,  see  Complex  quantities. 

addition,  113 

multiplication,  113 
Ventilating  ducts,  2.94 
Volt,  1 

Voltage  control  of  rotary  converters, 
385,  391 

ratios  of  rotary  converters,  383 

Wattless  component,  108,  110,  145 
Wattmeter,  146 

compensated,  148 

connections,  147 

correction  factor,  150 

errors,  150 
Wave  analysis,  196 

distortion  in  transformers,  189 
Waves,  generation  of  e.m.f.,  30 
Weight  of  transformers,  223 
WiNTER-EiCHBERG  motor,  407 

Zero  vector,  108 
"Zig-zag"  flux,  374 


A. TT  TNI**-**  i  i IRE  TO  nc.'*""- 


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